strength of materials and structures lecturer guide
TRANSCRIPT
STRENGTH OF MATERIALS AND STRUCTURES
N5Strength of Materials
and StructuresLecturer Guide
Henty Wickens
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© Future Managers 2018
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ISBN 978-0-6391-0078-4
First edition 2018
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Published byFuture Managers (Pty) LtdPO Box 13194, Mowbray, 7705Tel (021) 462 3572Fax (021) 462 3681E-mail: [email protected]: www.futuremanagers.com
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Contents
Module 1 Stress and strain and testing of materials 1
Module 2 Strain energy 9
Module 3 Compound bars and temperature-induced stresses 24
Module 4 Thin cylinders and riveted joints 44
Module 5 Loading of beams 59
Module 6 Simple bending of beams 80
Module 7 Columns and struts 105
Module 8 Shafts 113
Module 9 Forces 127
Module 10 Structural frameworks 135
Module 11 Concrete and foundations 148
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• Good knowledge of Engineering Science N4• Basic knowledge of calculating stress• Drawing of straight-line graphs• Know the conditions for equilibrium• Newton’s third rule
Learning outcomes
When students have completed this learning module, they should be able to:• Identify the diff erent stresses• Calculate direct and shear stress• Know what strain is and calculate it• Know what modulus of elasticity (Young’s modulus) for materials is and to
calculate it• Draw a stress-strain graph and use it to obtain information about a material• Describe diff erent tests on materials for diff erent strengths
Guidelines for students
• Make sure they understand the basic concept of the theory about stress and strain• Always work from basic principles and avoid formulas not directly connected to
fi rst principles• Always make sketches of the member or whatever is described in a problem• Read through the module and underline important facts• Make sure they know the answers to the self-check, before doing the exercises
Stress and strain and testing of materials1
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2 N5 Strength of materials and structures Lecturer Guide
Exercise 1.1 SB page 19
1. It is very important to sketch the problem.
1.1 Working stress = maximum stress __________ FOS
= 120 ___ 3
= 40 MPa
1.2 Safe load = σ.A
= 40 × π _ 4 (0,0094562)
= 2,809 kN
2. Effective area = (50 – 12) × 15
= 570 mm2
∴ σ = F __ A = 60k _______ 570 × 1 0 −6
= 105,26 MPa
3.
ø60 ø20
50 100
20 kN 20 kN
(1)
(2)
Basic equation: E = σ __ ε = F __ A __ x _ L = FL __ Ax = σ L __ x
3.1 Maximum stress smallest area = σ = F __ A = 20k ____ π _ 4 90 2 2
= 63,662 MPa
3.2 x T = x 1 + x 2
= F L 1 ___ A 1 E + F L 2 ___ A 2 E
= 20k ____ 200G [ 0,05 ____ π _ 4 0,0 6 2
+ 0,1 ____ π _ 4 0,0 2 2
] = 0,0336 mm
3.3 ε T = ε 1 + ε 2 ∴ ε 1 = σ 1 __ E = F ___ A 1 E = 20k ____
π _ 4 0,0 6 2 × 200G = 3,537 × 10–5
ε2 = σ 2 __ E = F ___ A 2 E = 20k ____
π _ 4 0,0 2 2 × 200G = 3,183 × 10–4
∴ ε T = ε 1 + ε 2 = 3,537 × 10–5 + 3,183 × 10–4 = 3,537 × 10−4
15
5060 kN
151250
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3Module 1 • Stress and strain and testing of materials
4. E = 200 GPa
ø60 ø80ø20F F
(1)
(2)
(3)
σmax = 100 MPa
4.1 Maximum stress smallest area ∴ F = σ max A 2
= 100 M × π _ 4 0,0 2 2 = 31,416 kN This is the max load which can be applied due to the weakest section.
4.2 σ1 = F __ A 1 = 31,416k _____
π _ 4 0,0 6 2 = 11,11 MPa
σ3 = F __ A 3 = 31,416k _____
π _ 4 0,0 8 2 = 6,25 MPa
5. σAL = 155 MPa
Eal = 69 GPa
σs = 465 MPa
Es = 207 GPa
σc = 247 MPa
EC = 110 GPa
5.1 Aluminium maximum allowable on working stress = σ max ___ FOS = 155 ___ 3 = 51,67 MPa
Steel maximum allowable on working stress = σ max ___ FOS = 465 ___ 3 = 155 MPa
Copper maximum allowable on working stress = σ max ___ FOS = 247 ___ 3 = 82,33 MPa
5.2 FAL = σAAA = 51,67 M × π _ 4 (0,052 – 0,042) = 36,523 kN
FS = σSAS = 155 M × π _ 4 (0,032 – 0,022) = 60,868 kN
FC = σcAc = 82,33 M × π _ 4 (0,092) = 523,76 kN
∴ Max force = 36,523 kN
The smallest load is the max because the other loads will damage the aluminium. The strain in aluminium will be higher as the max allowable for aluminium is 51,67 MPa.
5.3 σa = 51,67 MPa acting force
σs = F __ A = 36,523k __________ π _ 4 (0,0 3 2 − 0,0 2 2 )
= 93,01 MPa
σc = F __ A = 36,523k _____ π _ 4 0,0 9 2
= 5,74 MPa
This will be the stresses in the other materials when the max load is applied.
ø90ø40
80 100 60
OD = 50
L
OD = 30
IO = 20F F
AL
SteelCopperAL
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4 N5 Strength of materials and structures Lecturer Guide
5.4 Total extension: x T = x a + x s + x c
x s = σ s L s ___ E s = 93,01M × 0,1 _________ 207G = 0,0449 mm
x c = σ c L c ___ E c = 5,74M × 0,06 _________ 110G = 3,131 × 10–3 mm
From : x a = x T – ( x s + x c )
= 0,0981 – (0,0449 + 3,131 × 10–3)
= 0,050069
But x a = x hollow + x solid due to the two different areas.
= F L H ____ A H E a + F L s ___ A s E a
∴ 0,050069 × 10–3 = 36 523 _____ 69G [ L H __________
π _ 4 (0,0 5 2 − 0,0 4 2 ) + (0,08 − L H )
_______ π _ 4 0,0 5 2
] (∴ LS = (80 – LH))
∴ 94,5914 = 1 414,711 LH + 509,296(0,08 – LH)
53,848 = 905,415 LH
LH = 59,473 mm
6.
Force
Extension
6.1
6.2yield point
max load point
point of fracturelimit of proportionality
6.3
6.4
7. 70
ø15
7.1 σ = F __ A = 7k _____
π _ 4 0,01 5 2 = 39,612 MPa (see 3.1 on page 14 in the Student Book)
7.2 E = σL __ x = 39,612 M × 0,07 __________ 35 × 1 0 −6
= 79,224 GPa (see 3.2 on page 14 in the Student Book)
8. 8.1 σ = F __ A = 28k _____ π _ 4 0,01 2 2
= 247,57 MPa
8.2 E = σL __ x = 247,57 M × 0,06 __________ 0,19 × 1 0 −3
= 78,18 GPa
8.3 σmax = F max ____ Area = 52k _____ π _ 4 0,01 2 2
= 459,78 MPa
8.4 Actual stress = 40k _____ π _ 4 0,00 9 2
= 628,76 MPa
8.5 % elongation = x _ L × 100 ___ 1 = 8 __ 60 × 100 ___ 1 = 13,33%
8.6 % reduction in area = ( 1 2 2 − 9 2 ______ 1 2 2
) × 100 ___ 1 = 43,75%
Basic equation:E = σ _ ε = F/A ___ x/L = FL __ Ax = σL __ x (see pages 6 and 9 in the Student Book)
Basic equation:E = σ _ ε = F/A ___ x/L = FL __ Ax = σL __ x
x = 35 × 10–6 m F = 7 kN
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5Module 1 • Stress and strain and testing of materials
9.
Change units of extension to metres.
F(kN) 3,5 8,5 13,5 18,8 23,5 28,5 31 33,5 35 36
x × 10–3 9 22 35 48 61 74 87 93 104 157
9.1 E = FL __ Ax = 28,5k × 0,060 ____________ π _ 4 0,01 2 2 × 74 × 1 0 −6
= 204,32 GPa
9.2 σ = F __ A = 27k _____ π _ 4 0,01 2 2
= 238,73 MPa
9.3 σ = F __ A = 46k _____ π _ 4 0,01 2 2
= 406,73 MPa
9.4 % x = x _ L × 100 ___ 1 = 9 __ 60 × 100 ___ 1 = 15%
9.5 % A = 1 2 2 − 8 2 ______ 1 2 2
× 100 ___ 1 = 55,56%
10. 10.1
Forc
e kN
x × 10–3
0
5
10
20 40 60 80 100 120 140 160
15
20
25
30
35
40
–A27
70
σ M
Pa
Strain × 10–4
0
40
20
10 20 30 40 50 60
60
80
100
120
140
160
180
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6 N5 Strength of materials and structures Lecturer Guide
10.2 E = FL __ Ax = 42 × 1 0 3 ______ 6 × 1 0 −4
= 70 GPa
Study the graph and select any value on the straight line of the graph. Take values from the table given which is on the straight line (see page 8 in the Student Book).
10.3 σ = F __ A = 21k _______ 100 × 1 0 −6
= 210 MPa (read pages 7–8 in the Student Book)
11.
F 25 55 80 95 104 109 114 117 118 120x mm 0,08 0,176 0,255 0,303 0,332 0,35 0,41 0,52 0,88 1,75MPa σ 62,87 138,33 201,2 238,9 261,56 274,14 286,7 294,3 296,8 301,8
ε = x _ L 3,2 × 10–4
7,04 × 10–4
1,02 × 10–3
1,212 × 10–3
1,328 × 10–3
1,4 × 10–3
1,64 × 10–3
2,08 × 10–3
3,52 × 10–3
7 × 10–3
× 10–4 3,2 7,04 10,2 12,12 13,28 14 16,4 20,8 35,2 70
11.1 E = σ __ ε = 201,2 M _______ 1,02 × 1 0 −3
= ± 197,3 GPa 80 kN force on straight line ∴ use 201,2 MPa stress. If the stress for the
55 kN or 95 kN was used, the answer would be close to 197,3 GPa. 11.2 σ = F __ A = ± 261,56 MPa Read from graph where straight line stops. 11.3 ± 286,7 MPa Read from graph just after straight line stops.
12. Safe stress = 465 ___ 5 = 93 MPa See page 9 in Student Book.
OD = 100
ID = 220
Force in wall = σ.A = 93M × π _ 4 ( 0,2 2 2 − 0, 1 2 ) = 2 804,814 kN
σ M
Pa
Strain × 10–4
0
50
10 20 30 40 50 60 70
250
200
100
150
300
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7Module 1 • Stress and strain and testing of materials
∴ Internal pressure = F __ A = 2 804,814k _______ π _ 4 01 1 2
= 357,12 MPa
• Calculate the area and store in memory of calculator. ∴ σ = F __ A Divide each force by the area to obtain stress for each force. • To calculate the strain, use calculator and divide the extension by the gauge
length for each extention (both can be in mm not to change to m – keep the units the same, m or mm).
Exercise 1.2 SB page 24
1. σ = 80 MPa
ε = 1,054 × 10–3
L = 120 mm
1.1 E = σ __ ε = 80 M ________ 1,054 × 1 0 −3
= 75,9 GPa 1.2 ε = x _ L ∴ x = Lε = 120 × 1,054 × 10–3 = 0,1265 mm 1.3 σ = F __ A ∴ A = F _ σ = 24k ___ 80M = 3 × 10–4 m2
A = wt ∴ 3 × 10–4 = 0,05 × t ∴ t = 6 mm
2.
σmax = 100 MPa
xT = 0,0718 mm
E = 200 GPa 2.1 σ = F __ A ∴ A = F _ σ = 60k ____ 100M = 6 × 10–4 = π _ 4 D2
D = 27,64 mm
2.2 x T = x 1 + x 2 x 2 = x T – x 1 = 0,0718 – ( σL __ E ) = 0,0718 – ( 100M × 0,08 ________ 200G ) = 0,0718 – 0,04 = 0,0318 mm x 2 = FL ___ AE = 60k × 3L _________
π _ 4 0,0 6 2 × 200G = 0,0318 × 10–3
∴ L = 100 mm ∴ 2L = 200 mm
t
120
W
ø60 ø6060 kN 60 kN
(2)
L 8 2L
(1)
(2)
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8 N5 Strength of materials and structures Lecturer Guide
3. 3.1 Allowable stress = 84 __ 4 = 21 MPa See page 8 in the Student Book. Point C is the point that the particles in
the material separate before setting and the safe stress must be less. ∴ Use F.O.S.
3.2 F = σ.A = 21M × π _ 4 0,022 = 6,597 kN 3.3 x = σL __ E = 21M × 0,25 _______ 200G = 0,026 mm
Exercise 1.3 SB page 28
1. F = σ.A = 300 M × πD × t (use the shear area πDt) = 300 M × π × 0,01 × 0,02 = 188,496 kN
2. 2.1 OD = 110
ID = 90 Figure 1
F = σ.A = 60 M × π × 0,09 × 0,012 (shear area in collar = πdt) = 203,575 kN
2.2 σ = F __ A = 203,575 _____ π _ 4 0,0 9 2
= 32 MPa (shaft diameter)
2.3 σ = F __ A = 203,575k __________ π _ 4 (0,1 1 2 − 0,0 9 2 )
(contact area of collar – see Figure 1)
= 64,8 MPa
3. 3.1 Area: σ = F __ A ∴ F _ σ = 100k ____ 100M = 1 × 10–3 = π _ 4 D2
D = 35,68 mm
3.2 τ = F __ A = 100k _______ 2 × π _ 4 0,0 2 2
= 159,155 MPa (pin shear in two areas ∴ 2 × area of pin [cross-sectional area])
3.3
Figure 2
Two legs of fork ∴ Area = 2(80 – 20)15 = 1 800 mm2
σt = F __ A = 100k ________ 1 800 × 1 0 −6
= 55,56 MPa
15
80
8020
Remember answers for graph questions will always be ± values (questions 9–11).
In practice more tests are done to get the correct answers. We only do one calculation.
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MODULE
2 Strain energy
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this learning module:• Knowledge of the formulae used in Module 1 of this book• How to calculate work done• What potential energy is
Learning outcomes
When students have completed this learning module, they should know:• What a gradually applied load is• What a shock load is• What a suddenly applied load is• What strain energy is• What resilience is• How to calculate stresses in single and compound members due to the three ways
in which loads can be applied
Guidelines for students
• Make use of sketches when you do calculations and always work from fi rst principles
• Working from fi rst principles will give students more knowledge of the subject and they will also understand the theory of the subject better
• Make a good study of the work content and underline all important facts• Before starting with the exercise, make sure they understand the theory of the
module and know the answers of the self-check exercises
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10 N5 Strength of materials and structures Lecturer Guide
Exercise 2.1 SB page 46
1. U = 1 _ 2 Fx = 1 _ 2 F ( FL ___ AG ) = F 2 L ___ 2AG
= (60k ) 2 × 0,5 ___________ 2 × π _ 4 0,04 × 206G
= 3,477 J
2. 2.1
UT = U1 + U2 = 1 _ 2 F x 1 + 1 _ 2 F x 2
= 1 _ 2 F( x 1 + x 2 )
= 1 _ 2 F [ F L 1 ___ A 1 E + F L 2 ___ A 2 E
] = FL __ 2E [ L 1 __ A 1
+ L 2 __ A 2 ]
= ( 100k ) 2 ______ 2 × 206G [ 0,6 ____ 0,0 5 2
+ 1 ____ 0,0 8 2
] = 9,618 J
2.2 xT = x1 + x2
= F L 1 ___ A 1 E + F L 2 ___ A 2 E
= 100k ____ 206G [ 0,6 ____ 0,0 5 2
+ 1 ____ 0,0 8 2
] = 0,192 mm
2.3 εT = ε1 + ε2
= x 1 __ L 1 + x 2 __ L 2
= F L 1 ____ A 1 E L 1 + F L 2 ____ A 2 E L 2
= 100k ____ 206G [ 1 ____ 0,0 5 2
+ 1 ____ 0,0 8 2
] = 2,7 × 10–4
2.4 σmax = F __ A 1 = 100k ____
0,0 5 2 = 40 MPa (max stress in smallest area)
3. ID = 50
OD = 80
E = 200 GPa
U = 30 J
801 000
600 80
50
50
2
1
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11Module 2 • Strain energy
3.1 U = 1 _ 2 Fx
= 1 _ 2 F ( FL ___ AE ) ∴ 30 = F 2 L ___ 2AE = F 2 × 0,7 ________________
2 × π _ 4 (0,0 8 2 − 0,0 5 2 )200G
∴ F2 = 3,676 × 1 0 10 ________ 0,7
F = √ __________
5,251 × 1 0 10 = 229,149 kN
3.2 σ = F __ A = 229,149k __________ π _ 4 (0,0 8 2 − 0,0 5 2 )
= 74,811 MPa
3.3 x = σL __ E = 74,811 M × 0,7 _________ 200G = 0,262 mm
3.4 ε = σ _ E = 74,811M ______ 200G = 3,741 × 10–4
4.
E = 210 GPa
4.1 εT = ε1 + ε2 + ε3
= σ 1 __ E + σ 2 __ E + σ 3 __ E
= 1 _ E [ F __ A 1 + F __ A 2
+ F __ A 3 ]
= F _ E [ 1 __ A 1 + 1 __ A 2
+ 1 __ A 3 ]
∴ 7,548 ×10–4 = 200k ____ 210G [ 1 __ A 1 + 1 ____
π _ 4 0, 1 2 + 1 ____
0,0 5 2 ]
∴ 1 __ A 1 = 265,216
A1 = 3,771 × 10–3
∴ π _ 4 (0,0 8 2 − d 2 ) = 3,771 × 10–3
d = 40 mm 4.2 xT = x1 + x2 + x3
= F L 1 ___ A 1 E + F L 2 ___ A 2 E
+ F L 3 ___ A 3 E
= F _ E [ L 1 __ A 1 + L 2 __ A 2
+ L 3 __ A 3 ]
= 200k ____ 210G [ 0,6 __________ π _ 4 (0,0 8 2 − 0,0 4 2 )
+ 0,2 ____ π _ 4 0, 1 2
+ 0,5 ____ 0,0 5 2
] = 0,366 mm 4.3 Check for smallest area: A1 = π _ 4 (0,0 8 2 − 0,0 4 2 ) = 3,77 × 10–3 m2
A3 = 0,052 = 2,5 ×10–3 m2
600200
50 × 50OD = 80 ø0 = 200
ID = ?
500
200 kN 200 kN1
2 3
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12 N5 Strength of materials and structures Lecturer Guide
σmax = F ___ A min
= 200k _______ 2,5 × 1 0 −3
= 80 MPa
We must check for the smallest area to calculate because the hollow area can be smaller, depending on diameters.
4.4 UT = 1 _ 2 F x T = 1 _ 2 × 200k × 0,366 × 10–3
= 36,6 J
5.
ECi = 80 GPa EC = 100 GPa
5.1 UT = U1 + U2
= 1 _ 2 F x 1 + 1 _ 2 F x 2 = 1 _ 2 F [ x 1 + x 2 ]
∴ = F _ 2 [ F L 1 ___ A 1 E 1 + F L 2 ___ A 2 E 2
] = F 2 __ 2 [ L 1 ___ A 1 E 1
+ L 2 ___ A 2 E 2 ]
= (250k ) 2 _____ 2 [ 2,5 ________ π _ 4 0, 1 2 × 80G
+ 2,5 ________ π _ 4 0, 1 2 × 100G
] = 223,812 J
5.2 UT = 1 _ 2 F x T
= 1 _ 2 F [ FL ___ AE ] ∴ 223,812 = F 2 L ___ 2AE = (250k ) 2 × 5 _______ 2A × 210G
∴ A = 3,324 × 10–3
∴ π _ 4 d 2 = 3,324 × 10–3
d = 65,06 mm
Exercise 2.2 SB page 57
E = σ __ ε = F/A ___ x/L = FL __ Ax = σL __ x (basic equation still applies)
d = 80
E = 206 GPa
W = 4 kN
h = 200 mm
ø100 Ci ø100 C
2,5 m
250 250
2,5 m
d = ?
Students must show more calculation steps.
L = 4 m
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13Module 2 • Strain energy
1. 1.1 PE = U
W(h + xT) = 1 _ 2 F x T ( x T = FL ___ AE ) ∴ 4k ( 0,2 + F × 4 _________
π _ 4 0,0 8 2 × 200G ) = F 2 × 4 ___________
2 × π _ 4 0,0 8 2 × 206G
∴ 4k(0,2 + 3,863 × 10–9 F) = 1,931 × 10–9 F2
∴ 800 + 1,545 × 10–5 F = 1,931 × 10–9 F2
÷ 1,931 × 10–9: ∴ F2 – 8 002,072 F – 4,143 × 1011 = 0
F = −(−8 002,072 ± √ ______________________________
(−8 002,072 ) 2 − 4(1)(−4,143 × 1 0 11 ) _________________________________ 2 × 1
F = 8 002,072 ± 1 287 337 ______________ 2
= 647,67 kN
∴ σ = F __ A = 647,67k _____ π _ 4 0,0 8 2
= 128,85 MPa
1.2 PE = U W(h + x) = 1 _ 2 Fx ∴ 40k(0,02 + 3,863 × 10–9 F) = 1,931 × 10–9 F2
∴ 800 + 1,545 × 10–4 F = 1,931 × 10–9 F2
÷ 1,931 × 10–9: ∴ F2 – 80 010,357 F – 4,143 × 1011
F = 80 010,357 ± √ _______
b 2 − 4ac ______________ 2(1)
F = 80 010,357 ± 1 289 796,1 _______________ 2 = 684,9 kN
σ = F __ A = 684,9 ____ π _ 4 0,0 8 2
= 136,2 MPa
2. E = 210 GPa
W = 200 × 9,81 = 1 962 N
2.1 PE = U
W(h + xT) = 1 _ 2 F x T
xT = x1 + x2
= F L 1 ___ A 1 E + F L 2 ___ A 2 E
= F ____ 210G [ 1 ____ 0,0 4 2
+ O 1 8 ____ π _ 4 0,0 2 2
] xT = 1,51 × 10–8 F
1 40 × 46
2L = 0,8 m d = 20 h = 300
1 m
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14 N5 Strength of materials and structures Lecturer Guide
Substitute in :
∴ 1 962(0,3 + 1,51 × 10–8 F) = 1 _ 2 F × 1,51 × 1 0 −8 F
∴ 588,6 + 2,963 × 10–5 F = 7,55 × 10–9 F2
÷ 7,55 × 10–9: ∴ F2 – 3 924,503 F – 7,796 × 10–10 = 0
F = − (−3 924,503) ± √ _______
b 2 − 4ac _________________ 2(1)
F = 3 924,503 ± 558 441,1 ______________ 2
= 281,183 kN
A1 = 0,062 = 1,6 × 10–3 m
A2 = π _ 4 0,0 2 2 = 3,162 × 10–4
∴ σmax = 281,183k ________ 3,142 × 1 0 −4
= 894,92 MPa
2.2 From xT = 1,51 × 10–8 F
= 1,51 × 10–8 × 281,183 k
= 4,246 mm
2.3 UT = 1 _ 2 F x T = 1 _ 2 × 281,183 × 4,246 × 10–3
= 596,932 J
2.4 σ1 = W __ A 1 = 1 962 ____
0,0 4 2 = 1,226 MPa
σ2 = W __ A 2 = 1 962 ________
3,1 4 2 × 1 0 −4 = 6,244 MPa
2.5 U1 = 1 _ 2 F x 1 = F 2 L ___ 2AE = 1 96 2 2 × 1 ___________ 2 × 0,0 4 2 × 210G
= 5,728 × 10–3 J
U2 = F 2 L ___ 2AE = 1 96 2 2 × 0,8 ___________ 2 × π _ 4 0,0 2 2 × 210G
= 0,0233 J
UT = U1 + U2 = 0,0291 J Final strain is the strain under load gradually applied.
3. A = 700 mm2
xT = 2,5 mm
E = 206 GPa
PE = U
3.1 E = σL __ x
∴ 206G = σ3,5 _______ 2,5 × 1 0 −3
σ = 147,43 MPa
3.2 PE = U ∴ W(h + x T ) = 1 _ 2 F x T
L = 3,5 mh = 15 mm
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15Module 2 • Strain energy
∴ W(0,015 + 2,5 × 10–3) = 1 _ 2 σ.A × 2,5 × 10–3
0,0175W = 1 _ 2 × 147,143M × 700 × 10–6 × 2,5 × 10–3
W = 7,357 kN
3.3 Mass = weight _____ g = 7,357k _____ 9,81
= 749,96 kg
4. 4.1 F = 2W
∴ F = 2 × 50 = 100k
∴ σ = F __ A = 100k _____ π _ 4 0,02 6 2
= 188,349 MPa
4.2 xT = σL __ E = 188,349M × 3 _________ 200G = 2,825 mm
5.
5.1 σA = 50k ____ π _ 4 0,0 5 2
= 25,46 MPa
σB = 50k ____ π _ 4 0,0 3 2
= 70,74 MPa
σC = F __ A = 50k ____ π _ 4 0,0 7 2
= 12,99 MPa
5.2 xA = σ A L A ____ E = 25,46M × 0,4 ________ 200G = 0,051 mm
xB = σ B L B ___ E = 71,74M × 0,03 _________ 200G = 0,106 mm
xC = σ C L C ___ E = 12,99M × 1 0 6 × 0,9 ____________ 200G = 0,058 mm
5.3 UA = 1 _ 2 F x A = 1 _ 2 × 50k × 0,051 × 1 0 −3 = 1,275 J
UB = 1 _ 2 F x B = 1 _ 2 × 50k × 0,106 × 1 0 −3 = 2,65 J
UC = 1 _ 2 F x C = 1 _ 2 × 50k × 0,058 × 1 0 −3 = 1,45 J
5.4 Strain
εA = σ A __ E = 25,46M _____ 200G = 1,273 × 10–4
εB = σ B __ E = x B
__ L B = 0,106 × 1 0 −3 ________ 0,3 = 3,53 × 10–4
εC = x C __ L C = 0,058 × 1 0 −3 ________ 0,9 = 6,44 × 10–5
5.5 UT = U1 + U2 + U3 = 5,375 J (add the values in 5.3)
OR
UT = 1 _ 2 F x T
= 1 _ 2 × 50k (0,051 + 0,106 + 0,058)10–3
= 5,375 J
ø50 ø30 ø70
400 300
50 kN 50 kN
900
A B C
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16 N5 Strength of materials and structures Lecturer Guide
6. 6.1 U = 1 _ 2 Fx = 1 _ 2 F FL ___ AE
∴ 3 = F 2 × 0,6 ___________ 2 × π _ 4 0,0 5 2 × 200G
F = 62,666 kN
6.2 σ = F __ A = 62,666k _____ π _ 4 0,0 5 2
= 31,92 MPa
7.
30 x 30 10 x 10
600 400
F F
7.1 xT = x1 + x2
= FL1 ___ A 1 E + F L 2 ___ A 2 E
∴ 0,15 × 10–3 = F ____ 209G [ 0,6 ____ 0,0 3 2
+ 0,4 ____ 0,0 1 2
] 0,15 × 10–3 = 2,2329 × 10–8 F F = 6,718 kN
7.2 σmax = F ___ A min = 6,718k _____ 0,0 1 2
(max stress in smallest area)
= 67,18 MPa
8.
L = 1,5 mø = 30
W = 4 kgx = 0,08 mm
E = FL __ Ax
= 4 × 9,81 × 1,5 _____________ π _ 4 0,0 3 2 × 0,08 × 1 0 −3
= 1,041 GPa
∴ PE = U
∴ W(h + xT) = 1 _ 2 F x T
∴ 4 × 9,81 ( 0,003 + FL ___ AE ) = 1 _ 2 F ( FL ___ AE ) ∴ 4 × 9,81 ( 0,003 + F × 1,5 ___________
π _ 4 0,0 3 2 × 1,041G ) = F 2 × 1,5 _____ 2AE
∴ 4 × 9,81(0,003 + 2,0385 × 10–6 F) = 1,019 × 10–6 F2
∴ 0,11772 + 7,999 × 10–5 F = 1,019 × 10–6 F2
÷ 1,019 × 10–6: ∴ F2 – 78,5 F – 115 525,02 = 0
F = 78,5 ± √ _______
b 2 − 4ac ___________ 2(1)
= 78,5 ± 684,297 _________ 2
= 381,4 N
∴ σ = F __ A = 381,4 ____ π _ 4 0,0 3 2
= 539,569 kPa
9. PE = U
∴ W(h + x) = 1 _ 2 Fx
25k ( h + σL __ E ) = 1 _ 2 σ.A × σL __ E
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17Module 2 • Strain energy
25k ( h + 102M × 0,8 _______ 205G ) = (102M)2 ______ 2 × π _ 4 0,1 2 2 × 0,8 ____ 205G
∴ 25k (h + 3,98 × 10–4) = 229,593
h = 8,786 mm
10. 10.1 U = 1 _ 2 F x T = 1 _ 2 σ.A. σL __ E = σ 2 AL ____ 2E
8 = σ 2 × 0,0 3 2 × 2 _________ 2 × 198G
σ2 = 1,76 × 1015
σ = 41,95 MPa
10.2 10.2.1 U = 1 _ 2 Fx ∴ 8 = F 2 L ___ 2AE = F 2 × 2 ___________ 2 × 0,0 3 2 × 198G
∴ F = 37,757 kW
10.2.2 U = Wx = W(2W)L ______ EA = 2WL ____ EA (F = 2W for suddenly applied)
8 = 2 × W 2 × 2 ________ 0,0 3 2 × 198G
W = 18,879 kN
10.2.3 W(h + x) = U
W(0,02 + xT) = 8
But xT ∴ 8 = 1 _ 2 F x T
∴ 8 = 1 _ 2 × 37,757 k x T
∴ xT = 4,238 × 10–4
Substitute in ∴ W(0,02 × 4,238 × 10–4) = 8
W = 391,7 N11. 11.1 U = 1 _ 2 Fx
= 1 _ 2 × 2W × 2WL ____ AE (F = 2W)
∴ 8 = 2 W 2 L ____ AE
= 2 W 2 × 1,5 ____________ 320 × 1 0 −6 × 200G
∴ W = 13,064 kN
∴ Force in rod = F = 2W = 26,128 kN
11.2 U = 1 _ 2 Fx (F = W)
8 = F 2 L ___ 2AE
= F 2 × 1,5 __________ 2 × 320 × 1 0 −6 E
(E = 200G)
∴ F = 26,128 kN
IMPORTANT For U to be constant the force F must also be constant.∴ From 10.2.1 F = 37,757 kxT
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18 N5 Strength of materials and structures Lecturer Guide
11.3 PE = U
∴ W(h + x) = 1 _ 2 Fx
W(h + x) = F 2 L ___ 2AE = 8
∴ F 2 × 1,5 _____ 2AE = 8
F = 26,128 kN
11.4 11.4.1 F = 2W x = FL ___ AE
∴ = 26,128k × 1,5 _________ AE = 0,612 mm
11.4.2 x = FL ___ AE = 26,128k × 1,5 _________ AE = 0,612 mm
11.4.3 x = W(h + x) = 1 _ 2 Fx
W(h + x) = 8 = 1 _ 2 Fx
= 1 _ 2 × 26,128k
∴ x = 0,612 mm
12. xT = 1 mm
W = 300 × 9,81
E = 210 GPa
PE = U W(h + xT) = 1 _ 2 F x T
xT = x1 + x2
= F _ E [ L 1 __ A 1 + L 2 __ A 2
] 1 × 10–3 = F ____ 210G [ 0,8 _______
350 × 1 0 −6 + 0,6 _______
400 × 1 0 −6 ]
∴ F = 55,472 kN
Substitute in : 300 × 9,87(h + 1 × 10–3) = 1 _ 2 55,472k × 1 × 10–3
h = 8,424 mm
13. 13.1 U = 1 _ 2 Fx 4 = 1 _ 2 F × FL ___ AE = F 2 × 2 _______________
2 × 2 500 × 1 0 −6 × 200G
∴ F = 44,721 kN
13.2.1 F = 2W = 2 × 44,721 = 89,442 kN
IMPORTANT Students must show all values.
IMPORTANT Students must show all values. This is only showing the metod of calculation.
L = 600 h = ?400 mm2
L = 800 350 mm2
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19Module 2 • Strain energy
13.2 13.2.2 PE = ∪ W(h + xT) = 1 _ 2 Fx
44,721 ( 5 × 1 0 −3 + F × 2 _________ 2 500 × 1 0 −6 E
) = F L 2 ___ 2AE ∴ 223,605 + 1,789 × 10–4 F = 2 × 10–9 F2 ÷ 2 × 10–9: ∴ F2 – 89 450F – 1,118 × 1011 = 0
∴ F = 89 450 ± √ _______
b 2 − 4ac ____________ 2(1)
= 89 450 ± 674 693,49 _____________ 2 = 382,072 kN
14. 14.1 F Fø50 ø30
60100
U = 10 J
UT = U1 + U2
∴ 10 = 1 _ 2 F( x 1 + x 2 ) x = FL ___ AE
∴ 10 = F 2 __ 2E [ 0,1 ____ π _ 4 0,0 5 2
+ 0,06 ____ π _ 4 0,0 3 2
] ∴ 10 × 2 × 200G = F2 [50,93 + 84,883]
F = 171,617 kN
14.2 σ1 = 171,617k ______ π _ 4 0,0 5 2
= 87,4 MPa
σ2 = 171,617k ______ π _ 4 0,0 3 2
= 242,79 MPa
14.3 x1 = σ 1 L 1 ___ E = 87,4M × 0,1 ________ 200G = 0,044 mm
x2 = σ 2 L 2 ___ E = 242,79M × 0,06 __________ 200G = 0,0728 mm
15.
ø25
ø12,5
600 = L
600 = L
If stress = 220 MPa when load falls h metres, stress is maximum in smallest area.
∴ Force in rod will be:
F = σ.A = 220 M × π _ 4 0,01252
= 26,998 kN
∴ xT = x1 + x2 = F L 1 ___ A 1 E + F L 2 ___ A 2 E
∴ xT = 26 998 × 0,6 __________ π _ 4 0,02 5 2 × 207G
+ 26 998 × 0,6 ________ π _ 4 0,012 5 2 × E
(E = 207G)
= 7,971 × 10–4 m
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20 N5 Strength of materials and structures Lecturer Guide
But: PE = U
W(h + xT) = 1 _ 2 F x T
∴ 12 × 9,81(h + 7,971 × 10–4) = 1 _ 2 × 26 998 × 7,971 × 10–4
h = 90,6 mm
16. 16.1
ø40 ø30 ø20
10060 80
E = 210 GPa
xT = x1 + x2 + x3 (x = FL ___ AE )
∴ 0,177 × 10–3 = F _ E [ 0,1 ____ π _ 4 0,0 4 2
+ 0,06 ____ π _ 4 0,0 3 2
+ 0,08 ____ π _ 4 0,0 2 2
] ∴ F = 88,68 kN
16.2 D = 1,2d
xT = FL ___ AE
∴ A = 70k × 924 _____________ 0,177 × 1 0 −3 × 210G
= 451,9774 mm2
A = π _ 4 ( D 2 − d 2 ) = 451,9774
D2 – d 2 = 575,4755
Substitute in : (1,2d)2 – d 2 = 575,4755
∴ 1,44d 2 – d
2 = 575,4755
d = 36,165 mm
D = 43,4 mm
17. 80 kN 80 kN
2 000 mm2
1 m 2 m
1 000 mm2
E = 200 GPa
UT = U1 + U2
= F L 1 ___ A 1 E + F L 2 ___ A 2 E
= 80k ____ 200G [ 1 ________ 2 000 × 1 0 −6
+ 2 ________ 1 000 × 1 0 −6
] = 40 J
Uniform bar
Volume of step bar = V1 + V2 = A1L1 + A2L2
= (2 000 × 1 000) + (1 000 × 2 000)
= 4 × 106 mm3
Volume of step bar = Volume single bar
IMPORTANT Students must show all steps.
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21Module 2 • Strain energy
∴ 4 × 106 = AS × 3 000
AS = 1 333,33 m2
∴ U single = 1 _ 2 F x T
= F 2 L ___ 2AE
= (80k ) 2 × 3 _________________ 2 × 1 333,33 × 1 0 −6 × 200G
= 36 J
∴ 2 sections more energy.
18.
18.1 Max stress ⇒ small area ∴ F = σ.A = 200 m × 0,012
= 20 kN 18.2 xT = x1 + x2
But U = 1 _ 2 F x T
∴ xT = 2U __ F
= 2 × 1,3 _____ 20k
= 0,13 mm 18.3 Stress in round bar = F __ A 2
= 20k _____ π _ 4 0,02 5 2
= 40,744 MPa
xT = x1 + x2
= σ 1 L 1 ___ E + σ 2 L 2 ___ E
0,13 × 10–3 = 200M L 1 ______ 200G + 40,744M (0,25 − L 1 ) _____________ 200G
∴ × 200G: ∴ 26 × 106 = 200M L1 + 40,744M(0,25 – L1)
÷ 106: ∴ 26 = 200L1 + 10,186 – 40,744L1
15,814 = 159,256L1
L1 = 99,3 mm
10 × 10
ø25
250 – L 1
L 1
250
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22 N5 Strength of materials and structures Lecturer Guide
19. 19.1
S C S
d = 20
OD = 40ID = 35 Steel
W = ?
L = 400
h = 80
xT = 0,16 mm
xcT = xsT
∴ Use values of copper: ∴ σc ÷ EC = σ C L ___ x c
∴ σc = E C x c ___ L
∴ σc = 100G × 0,16 × 1 0 −3 ____________ 0,4
= 40 MPa
∴ σ Steel = E S x S ___ L
= 198G × 0,16 × 1 0 −3 ____________ 0,4
= 79,2 MPa
19.2 UT = UC + Usteel
∴ UC = 1 _ 2 Fx
= 1 _ 2 σ.Ax
= 1 _ 2 × 40M × π _ 4 0,0 2 2 × 0,16 × 1 0 −3
= 1,005 J
US = 1 _ 2 Fx
= 1 _ 2 σ.Ax
= 1 _ 2 × 79,2M × π _ 4 (0,0 4 2 − 0,03 5 2 ) × 0,16 × 1 0 −3
= 1,866 J
19.3 PE = U
W(h + x) = U1 + U2
W(0,08 + 0,16 × 10–3) = 1,005 + 1,866
0,08016W = 2,871
W = 35,82 N
(Can also use values of steel: LS = LC)
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23Module 2 • Strain energy
20.
σ = 80 mPa
d = 25
E = 200G
PE = U
W(h + x) = 1 _ 2 Fx
∴ x = σL __ E
= 80M × 1,5 _______ 200G
= 6 × 10–4 m
Sub. in : W(h + x) = 1 _ 2 σ.Ax
∴ W(0,15 + 6 × 10–4) = 1 _ 2 × 80M × π _ 4 0,2 5 2 × 6 × 1 0 −4
W = 78,227 N
Mass = 78,227 _____ 9,81
= 7,97 kg
h = 150 mm
L = 1,5 m
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MODULE
Compound bars and temperature-induced stresses3
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• Good knowledge of Modules 1 and 2• Ability to calculate for stress, strain and the change in length• Linear expansion of material subjected to change in temperature• Knowledge of screw thread terms• What a compound bar is• Nature of stresses due to cooling and an increase in temperature• Th e knowledge of Module 1• Must be able to use simultaneous equations• Ability to calculate moments of force
Learning outcomes
When students have completed this module they should be able to:• Calculate stresses in compound bars due to applied loads• Calculate the stresses in compound bars due to change in temperature• Calculate the resultant stresses in compound bars subjected to change in
temperature and an external force
Guidelines for the students
• To do the calculations in this module students must have a good knowledge of Module 1
• Th e same calculation principles are going to be used as they learnt in Module 1 and 2
• Th e knowledge will only be applied to a new theory for compound bars• Work from fi rst principles and do not rely on their memory and make sure they
understand the work from fi rst principles• Make use of sketches when they do calculations and always work from fi rst
principles• Working from fi rst principles will give students more knowledge of the subject
and they will also understand the theory of the subject better
IMPORTANT• Make a good study of the work content and underline all important facts• Before starting with the exercises make sure they understand the theory of the module and know the
answers of the self-check exercises
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25Module 3 • Compound bars and temperature-induced stresses
Exercise 3.1 SB page 73
1. C ø50S ø50200 kN 200 kN
1.1 FT = FC + FS
∴ 20k = σcAc + σsAs
= π _ 4 0, 0 2 ( σ c − σ s )
∴ 101,859 × 106 = σc + σs
xc = xs
∴ σ c L c ___ E c = σ s L s ___ E s
∴ σs = σ c E s ___ E c = σ c 207
____ 100 = 2,07σc
Substitute in ∴ 101,859 × 106 = 2,07σc
∴ σc = 33,179 MPa
σc = 68,68 MPa
1.2 Final length: LF = LO + x
= 100 + xc
∴ 0,100 + σ c L c ___ E c = 0,1 + 33,179 × 1 0 6 × 0,1 ___________ 100G
= 0,1 + 3,318 × 10–5
= 100,033 mm
2. S 200 mm2A 300 mm2 ε = 0,0005
ES = 209 GPa FT = FA + FS EA = 104,5 GPa
2.1 xs = xc
But x = σL __ E
∴σ = Ex __ L
= Eε ( x _ L = ε)
∴ σs = Esε and σA = Ecε
∴ σs = 209G × 0,0005 = 103,5 MPa
∴ σA = 104,5G × 0,0005 = 52,25 MPa
2.2 Fs = σsAs
= 103,5 × 106 × 200 × 10–6 = 20,7 kN
FA = σAAA
= 52,25M × 300 × 10–6 = 15,675 kN
Es = 207 GPaEc = 100 GPa
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26 N5 Strength of materials and structures Lecturer Guide
3. Copper
ø12 Steel
1505 5Copper
OD = 30
Id = 25
p = 1,5 mm pitch
ES = 200 GPa
EC = 100 GPa
3.1 FS = FC
∴ σcAs = σcAc
∴ σ s π _ 4 0,01 2 2 = σ c
π _ 4 (0,0 3 2 − 0,02 5 2 )
∴ σs = 1,91 σc
xT = xs + xc
xT = 2 _ 8 × 1,5 = 0,375 mm
∴ 0,375 × 10–3 = σ s L s ___ E s = σ c L c ___ E c
= σ s (150 + 10) ________ 200G + σ c (0,15)
_____ 100G
× 200G: ∴ 75 × 106 = 0,16 σs + 0,3 σc
Substitute in ∴ 75 × 106 = 0,16 (1,91σc) + 0,3 σc
= 0,6056 σc
∴ σc = 123,84 MPa
∴ σs = 236,54 MPa
3.2 xs = σ s L s ___ E s = 236,54M × 0,16 __________ 200G = 0,189 mm
xc = σ c L c ___ E c = 123,84M × 0,15 __________ 100G = 0,186 mm
4. OD = 500 200 kg x 9,81 N ES = 140G
Id = 420 EC = 14G
L = 2 m
4.1 FT = FI + FC
200 × 9,81 = σIAI + σcAc
= σ 1 π _ 4 ( 0, 5 2 − 0,4 2 2 ) + σ c ( π _ 4 0,4 2 2 )
1 962 = 0,058 σc1 + 0,139 σc
xS = xc
∴ σ s L s ___ E s = σ c L c ___ E c
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27Module 3 • Compound bars and temperature-induced stresses
∴ σS = σ c 140 ____ 14
= 10 σc
Substitute in ∴ 1 962 = 0,058(10 σc) + 0,139 σc
= 0,719 σc
∴ σc = 2,729 kPa
From σS = 27,29 kPa
4.2 xS = xc = σ c L c ___ E c = 2 729 × 2 ______ 14G = 3,9 × 10–4 mm
4.3 U = 1 _ 2 F x T = 1 _ 2 (200 × 9,81) × 3,9 × 10–7
= 3,826 × 10–4 J
5.
20ø
D = 100d = 80
D = 300
AT = π _ 4 0,32 = 0,07069 m2
AS = AP + AR
= π _ 4 ( 0, 1 2 − 0, 3 2 ) + ( π _ 4 × 0,0 2 2 × 6 ) = 2,827 × 10–3 + 1,885 × 10–3
AS = 4,712 × 10–3 m3
∴ AC = 0,07069 – 4,712 × 10–3 – π _ 4 0,082
AC = 0,061 m2
FT = FS + FC
300k = σsAs + σcAc
= σc4,712 × 10–3 + σc0,061
xs = xc
σ s __ E s = σ c __ G c
∴ σs = 15 σc
Substitute in ∴ 300k = 15 σc × 4,712 × 10–3 + σc 0,061
= 0,13168 σc
σc = 2,278 MPa
From σs = 34,17 MPa
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28 N5 Strength of materials and structures Lecturer Guide
6.
10 kN
Steelø4 ø6 ø4
1 2 3
ES = 200G
EC = 100G
x1 = x2 = x3
L1 = L2 = L3 = 3 m
No3 = d = 4
F = 30 kN
x = 60
L = 3
6.1 E3 = FL __ Ax
= 30k × 3 _________ π _ 4 0,00 4 2 × 0,06
∴ FT = F1 + F2 + F3
E3 = 119,37 GPa 10k = F1 + F2 + F3
x1 = x2 = x3
∴ x1 = x2
F 1 L 1 ___ A 1 E 1 = F 2 L 2 ___ A 2 E 2
∴ F 1 __________ π _ 4 0,00 4 2 × 200G
= F 2 __________ π _ 4 0,00 6 2 × 100G
∴ F1 = F 2 π _ 4 0,00 6 2 × 200G
___________ π _ 4 0,00 6 2 × 100G
F1 = 0,889 F2
x3 = x2
∴ F 3 L 3 ___ A 3 E 3 = F 2 L 2 ___ A 2 E 2
∴ F 3 ____________ π _ 4 0,00 4 2 × 119,37G
= F 2 __________ π _ 4 0,00 6 2 × 100G
∴ F3 = F 2 0,00 6 2 × 119,37G ____________
0,00 6 2 × 100G
= 0,531 F2
Substitute and in ∴ 10k = 0,889 F2 + F2 + 0,531 F2
∴ F2 = 4,132 kN
From ∴ F1 = 3,674 kN
From ∴ F3 = 2,194 kN
6.2 xT = x1 = σ 1 L 1 ___ E 1
= F 1 L 1 ___ A 1 E = 3,674k × 3 _________
π _ 4 0,00 4 2 200G
= 4,386 mm
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29Module 3 • Compound bars and temperature-induced stresses
7.
12
6 m8 m
8 kN
A = 900 m2
x1 = x2 E = 1 GPa
7.1 Load in each rope = 8 _ 2 = 4 kN
∴ σ = F __ A = 4k _______ 900 × 1 0 −6
= 4,44 MPa
7.2 Because it is a continuous rope.
∴ Average strength of rope = 6 + 8 ____ 2 = 7 m
∴ x = σ L AV ___ E
= 4,44m × 7 _______ 1G
= 31,08 mm
8.
10 N
9 m5 m
8.1 ∴ FT = F1 + F2
x1 = x2
∴ F 1 L 1 ____ A 1 G 1 = F 2 L 2 ____ A 2 G 2
[ A 1 = A 2 G 1 = G 2 ]
∴ × AG: ∴ F1L1 = F2L2
∴ F2 = F 1 9 ___ 5 = 1,8 F1
Substitute in ∴ 10k = F1 + 1,8 F1
∴ F1 = 3,571 kN ∴ F2 = 6,429 kN
σ1 = F 1 __ A = 3,571k _____ π _ 4 0,0 6 2
= 2,84 MPa
σ2 = F 2 __ A = 6,429k _____ π _ 4 0,0 4 2
= 5,12 MPa
8.2 ε1 = σ 1 __ E = 2,84M ____ 1,1G = 2,582 × 10–3
ε2 = σ 2 __ E = 5,12M ____ 1,1G = 4,655 × 10–3
8.3 x1 = x2 and ε2 = x 2 __ L 2
∴ x1 = ε1L1 ∴ x2 = ε2L2
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30 N5 Strength of materials and structures Lecturer Guide
x1 = 2,582 × 10–3 × 9 x2 = 4,655 × 10–3 × 5
= 23,24 mm = 23,28 mm
∴ U1 = 1 _ 2 F 1 x 1 = 1 _ 2 3,571k × 23,28 × 10–3
= 41,57 J
U2 = 1 _ 2 F 1 x 1 = 1 _ 2 × 6,429k × 23,28 ×10–3
= 74,83 J
8.4 3,571
6,419
RHx
10
(1 – x)
1
Moments about right ∴ 10x = 3,571 × 1
x = 0,3571 m
∴ 10 kN load is 357,1 mm from right
9. p = 1,2 mm
σB = 20 MPa
GS = 210 GPa
EB = 110 GPa
LB = 1,2 m
LT = 1,188 m
9.1 FB = FT xT = xB + xT
σBAB = σTAT
∴ 20 × π _ 4 0,01 5 2 = σT π _ 4 (0,02 4 2 − 0,01 6 2 )
σT = 14,063 MPa
9.2 FB = FT
σBAB = σTAT
σB = σT ( 0,024 2 − 0,016 2 __________ 0,015 2
) σB = 1,42 σT
xT = xB + xTU (xT = length × pitch)
(1,2 × 1,5)10–3 = σ B L B ___ E B + σ T L T
___ E T
1,8 × 10–3 = σ B 1,2 ____ 210G + σ T 1,188
_____ 110G
B
OD = 24 T
ø15
Id = 16
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31Module 3 • Compound bars and temperature-induced stresses
× 210G: ∴ 378 × 106 = 1,2σB + 2,268σT
Substitute in ∴ 378 × 106 = 1,2(1,42σT) + 2,268σT
∴ σbrass = 378M ____ 3072 = 95,17 MPa
σbolt = 135,14 MPa
Resultant stress = Stress tight nut + initial stress
∴ σbrass = 95,17 + 14,06 = 109,23 MPa
σsteel = 135,14 + 20 = 155,14 MPa
Exercise 3.2 SB page 92
IMPORTANT A sketch of the question must be made for the temperature question that will solve 90% of the problem.
1. L = 500L = 800
S Cø30 ø50 Δt = 90° – 20° = 70°
Free expansion: ∴ ∆LC = αctL = 18 × 10–6 × 70 × 0,8 = 1,008 mm
∴ ∆LS = αstL = 12 × 10–6 × 70 × 0,5 = 1,42 mm and LT = 1 300 mm ∴ LF = LT + ∆LC + ∆LS = 1 300 + 1,008 + 1,42 = 1 301,428 mm
2.
t = 50°
ε = 175 GPa
α = 15 × 10–6
2.1 With no free expansion: Stress = strain × E σ = εE
= αtL ___ L × E (∴ ε = change in lenghth ___________ original length = αtL ___ L = αt)
= αtE
= 15 × 10–6 × 50 × 175G
= 131,25 MPa
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32 N5 Strength of materials and structures Lecturer Guide
2.2 ε = αtL ___ L = αt = 15 × 10–6 × 50 = 7,5 × 1 0 −4
3.
C S
30
30 30
50
L = 400 t = 90°
Steel 30 × 50
Copper 30 × 30
3.1 FC = FS
δxc = δxs
∴ αtL – xc = αtL + xs
αctLc – αstLs = xc + xs
= F C L C ____ A C E C + F S L S ___ A S E S
÷ L: ∴ 90(17,5 – 13,5)10–6 = F ________ 0,0 3 2 × 100G
+ F ____________ 0,03 × 0,05 × 209G
× 209G: ∴ 75,24 × 106 = 2 322,22F + 666,66F
∴ 75,24 × 106 = 2 988,89F
∴ FS = 25,173 kN = FC
∴ σs = F C __ A C = 25,173k _______ 0,03 × 0,05 = 16,78 (T)
FC = 25,173k _____ 0,0 3 2
= 27,97 MPa (C)
xS
xC
C
400
S
αtLC
αtLS
δxC
δxS
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33Module 3 • Compound bars and temperature-induced stresses
3.2 t = T1 – T2 = 20 – (–5) = 25°C
C
S
L = 400αtLS
αtLC
xS
xC
δxC = δxS
FS = FC
∴ δxs = dxc
αtLs + xs = αtLc + xc
∴ αtLc – αtLs = xs + xc
÷L: ∴ LS = LC ∴ 25(17,5 – 13,5)10–6 = F C ___ A S E S + F C
____ A C E C
= F _________ (30 × 50)209G + F _________ (30 × 30)100G
× 209G: 20,9 × 106 = 666,67F + 2 322,22F
F = 6,993 kN ∴ σs = F __ A S
= 6 993 _____ 30 × 50 = 4,66 MPa (C)
∴ σc = F __ A C = 6 993 ____ ( 3 0 2 ) = 7,77 MPa (T)
3.3 εc = σ c __ E = 27,97M _____ 100G = 2,797 × 10–4
εs = σ s __ E = 16,78M _____ 209G = 8,029 × 10–5
4.
xC xS
αtLS
ø80 ø60
LS = 400LC = 300
S
δx
CαtLC
4.1 FC = FS
δxc = δxs
∴ αtLc – xc = xs – αtLs αctL + αstL = xs + xc
(18 × 10–6 × 80 × 0,3) + (12 × 10–6 × 80 × 0,4) (t = 80°) = F × 0,4 _________
π _ 4 0,0 8 2 × 100G + F × 0,3 _________
π _ 4 0,0 6 2 × 100G
× 209G: 171,36M = 79,58F + 222,82F = 302,39F
IMPORTANT Students must show all calculations.
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34 N5 Strength of materials and structures Lecturer Guide
∴ F = 566,68 kN
∴ σs = F __ A S = 566,68 × 1 0 3 ________
π _ 4 0,0 8 2 = 112,74 MPa
∴ σc = F __ A C = 566,68 × 1 0 3 ________ π _ 4 0,0 6 2
= 200,42 MPa
4.2 Use info for steel or use info for copper
∴ δxs = x – αtL
= σsL ___ E – αtL
= 112,74M × 0,4 _________ 210G – (12 × 10–5 × 80 × 0,4)
= 0,169 mm (shorter)
4.3 εs = σ s __ E s = 112,74M ______ 210G = 5,369 × 10–4
εc = σ c __ E c = 200,42M ______ 100G = 2,004 × 10–3
4.4 U = 1 _ 2 Fx = 1 _ 2 × 566,68k × 0,169 × 10–3
= 47,88 J
5.
xSxC
C
S
αtLC
αtLS
δx
5.1 δxc = δxs
αtLc – xc = αtLs + xs
αctL – αstL = xs + xc
(18 × 10–6 × 30 – 12 × 10–6 × 30) = F ____________ 400 × 1 0 −6 × 200G
+ F ____________ 400 × 1 0 −6 × 100G
× 200G: ∴ 36M = 2 500F + 5 000F ∴ F = 4,8 kN
σS = F __ A = 4,8k _______ 400 × 1 0 −6
= 12 MPa (C)
σc = F __ A = 4,8k _______ 400 × 1 0 −6
= 12 MPa (T)
Compound bar FT = FC + FS
100k = FC + FS
xs = xc
FL ___ AE = FL ___ AE
F S ____ 200G = F C ____ 100G
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35Module 3 • Compound bars and temperature-induced stresses
FS = 2FC
Substitute in ∴ 100k = FC + 2FC
∴ FC = 33,333 kN
FS = 66,667 kN
∴ σc = 33,333k _____ 400 = 83,33 MPa (T)
σs = 66,667k _____ 400 = 166,67 MPa (T)
∴ Rs σc = –12 – 83,33 = 95,33 MPa (T)
Rs σs = +12 – 166,67 = 154,67 MPa (T)
5.2 Use info of copper for cooling: δxc = αtL – xc
= 18 × 10–6 × 30 × 0,6 – FL ___ AE
= 3,24 × 10–4 – 4,8k × 0,6 ____________ 400 × 1 0 −6 × 100G
= 3,24 × 10–4 – 7,2 × 10–5
= 2,52 × 10–4 m = 0,252 mm ∴ LF = LO – δx = 600 – 0,252 = 599,748 mm
For tension of 100 kN, use steel:
xS = xC
∴ xS = σSLF ___ ES
= 166,67M × 0,599748 _____________ 200G
= 0,5 mm
∴ Final total length after coding and load:
LFT = FF + xload
= 599,748 + 0,5
= 600,248 mm
6.
xS
xC
S 300 mm2
C 400 mm2
400
αtLC
αtLS
δx
For Question 5.2 info of steel can also be used
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36 N5 Strength of materials and structures Lecturer Guide
6.1 δxc = δxs
αtL – x = αtL + x
αctL – αstL = xs + xc
∴ 40(18 –12)10–6 = F ________ 300 × 200G + F ________ 400 × 100G
× 200G: ∴ 48m = 3 333,31F + 5 000F
∴ F = 5 760 N
∴ σs = F S __ A S = 5 760 ______ 300 − m = 19,2 MPa (T)
σc = F S __ A S = 5 760 ______ 400 − m = 14,4 MPa (C)
6.2 FT = FS + FC
xs = xc
F S L ___ A S E S = F C L
____ A C E C
∴ FS = F C 300 × 200 ________ 400 × 100
= 1,5 FC
Substitute in ∴ 80 = 1,5 FC + FC
∴ FC = 32 kN ∴ FS = 48 kN
∴ σS = F S __ A S = 48k _______ 300 × 10–6 = 160 MPa (C)
σC = F C __ A C = 32k _______ 400 × 10–6 = 80 MPa (C)
6.3 Res: σc = +80 + 14,4 = 94,6 MPa (C)
Res: σs = +160 – 19,2 = 140,8 MPa (C)
6.4 LF = LO + δxc – xc (use info of one material)
Copper info: 400 + [αtL – x] – σ c L ___ E C
= 400 + [ 18 × 1 0 −6 × 40 × 94 − 14,4M × 0,4 ________ 100G ] – ( 80M × 0,4 _______ 100G ) = 400 + 0,23 – 0,32 = 399,91 mm
7. D = 40
d = 30
FC = FS
xT = xS + xC
Subjected to load means it is compressed.
1,5 m
ø12
20 20
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37Module 3 • Compound bars and temperature-induced stresses
7.1 Stress due to nuts tightening Total change in length:
xT = 80 ___ 360 × 1,75 = 0,389 mm
∴ xT = xs + xc
∴ 0,389 × 10–3 = F S L S ___ A S E S + F C L C
____ A C E C
= F 1,54 ________ π _ 4 0,01 2 2 200G
+ F 1,5 _____________ π _ 4 ( 0,0 4 2 − 0,0 3 2 ) 100G
× 200G: ∴ 77,8M = 13 616,59F + 5 456,74F ∴ F = 4,079 kN
∴ σs = F __ A S = 4,079k _____
π _ 4 0,01 2 2 = –36,07 MPa (T)
σc = F __ A C = 4,079k __________ π _ 4 (0,0 4 2 − 0,0 3 2 )
= +7,42 MPa (C)
Stress due to temperature cooling
C
C
S
20 20
xS
xC
αtLC
αtLS
δx
LC = 1,5
LS = 1,5 + 0,04
= 1,54
Cooling: FS = FC ∴ δxc = δxs
αtLc – xc = αtLs + xss
αctL – αstL = xs + xc [ ∵ x − − FL ___ AE ] ∴ (18 × 10–6 × 20 × 1,5) – (12 × 10–6 × 20 × 1,54)
= F1,54 ________ π _ 4 1 2 2 × 200G
+ F1,5 _____________ π _ 4 (4 0 2 − 3 0 2 ) × 100G
× 200G: ∴ 34,08M = 13 616,59F + 5 456,74F ∴ F = 1,788 kN
∴ σs = F __ A S = 1,788k _____
π _ 4 1 2 2 = 15,81 MPa (C)
σs = F __ A C = 1,788k ________ π _ 4 (4 0 2 − 3 0 2 )
= 3,25 MPa (T)
∴ Res: σc = +7,42 – 3,25 = +4,17 MPa (C)
Res: σs = –36,07 + 15,81 = –20,26 MPa (T)
7.2 Steel info
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38 N5 Strength of materials and structures Lecturer Guide
xF = xnut – δxs
δxs = δtL + x = (12 × 10–6 × 20 × 1,54) + 20,26M × 1,54 _________ 200G = 0,526 mm (cooled)
7.3 After nut was turned 80°:
xs = F S L S ___ A S E S = 4,079k × 1,54 _________ A E S
= 0,278 mm ∴ xc = xT – xs
= 0,389 – 0,278 = 0,111 mm Us = 1 _ 2 F x s
= 1 _ 2 × 4,079k × 0,278 × 10–3
= 0,567 J
Uc = 1 _ 2 F x c
= 1 _ 2 × 4,079k × 0,111 × 10–3
= 0,226 J
8.
3 000 kN
Sø3
1
Cø6
2
Sø3
3
8.1 FT = F1 + F2 + F3
FT = F1 + F2 + F3
∴ 3 000k
x1 = x2 = x3
∴ x1 = x2
∴ F 1 L 1 ___ A 1 E 1 = F 2 L 2 ___ A 2 E 2
(÷ L as L1 = L2)
∴ F1 = F 2 A 1 E 1 _____ A 2 E 2
= 3 2 200G _____ 6 2 100G
F2
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39Module 3 • Compound bars and temperature-induced stresses
F1 = 0,5 F2
x1 = x3
∴ F 1 ___ A 1 E 1 = F 3 ___ A 1 E 1
∴ F1 = F3 = 0,5 F2
Substitute and in : ∴ 3k = 0,5 F2 + F2 + 0,5 F2
= 2F2
∴ F2 = 1 500 N ∴ FC = 1,5 kN FS = 0,5 × 1,5 = 750 kN/wire
8.2 ∴ σc = 1 500 _____ π _ 4 0,00 6 2
= –53,05 MPa (T)
σs = 750 _____ π _ 4 0,00 3 2
= –106,1 MPa
FC = FS
Temp. δxc = δxs
αctL – αstL = xs + xc ( x = FL ___ AE ) 100(18 – 12)10–6 = F ___ A S E S
+ F ____ A C E C
∴ 6 × 10–4 = F ________ 2 × π _ 4 3 2 200G
+ F ______ π _ 4 6 2 100G
× 200G: ∴ 120M = 141 471,06F
∴ F = 848,23 N
∴ σc = 848,23 _____ 2 × π _ 4 3 2
= –60 MPa
σs = 848,3 ____ π _ 4 6 2
= –30 MPa
σres σc = –53,05 + 30 = –23,05 MPa (T)
Res σs = –106 – 60 = –166 MPa (T)
9. d = 20
D = 28
Ltube = 600 Lbolt = 620
σB = 30 MPa σT = 127,3 MPa
E = 200G 12 × 10–6
600
ø18
10 10
LB = 620
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40 N5 Strength of materials and structures Lecturer Guide
E = 100G 12 × 10–6
t = (30° – t2)
Stress in bolt and tube will be zero when their final lengths are equal.
Also
Res σB = 0 = –30 nut + 30 temp
σT = 0 = + 27,3 nut – 27,3 temp
But δxB = δxT
∴ αTtLT – αBtLB = xB + xT
∴ t[18 × 10–6 × 0,6 – 12 × 10–6 × 0,620] = σ B L B ___ E B + σ T L T
___ E T
∴ 3,36 × 10–6t = 30M × 0,62 _______ 200G + 27,3 × 0,6 ______ 100G
= 9,3 × 10–5 + 1,638 × 10–4
= 2,568 × 10–4
t = 76,43°
Tube compressive and bolt tensile when nut was tightened:
t = t1 – t2
76,43 = 30 – t2
∴ t2 = 30 – 76,43
= –46,43 °C
10.
800
800, 288
C
S
t1 = 20 °C
10.1 Bar must be cooled for both to have the same length.
αtL
αtL
0,288
C
S
∴ αctL = αstL + 0,288 × 10–3
∴ t(αcLc = αsLs) = 0,288 × 10–3
∴ t(18 × 10–6 × 0,800288 – 12 × 10–6 × 0,8) = 0,288 × 10–3
Tube length = 600Bolt length = 600 + 10 + 10 = 620
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41Module 3 • Compound bars and temperature-induced stresses
∴ t = 0,288 × 1 0 −3 __________ 4,805184 × 1 0 −6
= 59,94 °C t = t1 – t2
∴ t2 = t1 – t
= 20 – 59,94
= –39,94°C
10.2 αtLs = 12 × 10–6 × 0,8 × 59,96
= 0,575424 mm
∴ LN = 800 – 0,575424
= 799,425 mm
11.
Sø50
LS = 250 LC = 200t = 80 – 20 = 60°
Final
Cø50
αtLC
αtLS
xS
xC
δx
11.1 δtLc = 20 × 10–6 × 60 × 0,2 = 0,24 mm
δtLs = 12 × 10–6 × 60 × 0,25 = 0,18 mm
11.2 δxc = δxs
αtLc – xc = xs – αtLs
αtLc + αtLs = xs + xc
(From 11.1)
(0,24 + 0,18)10–3 = σ S L S ___ E S + σ C L C
___ E C
= σ S 0,25 ____ 200G + σ C 0,2
____ 95G
× 200G: ∴ 84M = 0,25 σs + 0,421 σc
But FC = FS
∴ σsAs = σcAc (AS = AC)
∴ σs = σc
Substitute in : ∴ 84M = 0,25 σc + 0,421 σc
∴ σc = σs = 125,19 MPa
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42 N5 Strength of materials and structures Lecturer Guide
12.
C
800
ø30
C
S
xC
xS
αtLC
αtLS
δx
OD 45ID 32
Final 12.1 FS = FC
δxc = δxs
αtLc + xc = αtLs + xs
αtLc – αtLs = xs + xc ( x = FL ___ AE ) 80(18 – 12)10–6 = F _______
π _ 4 3 0 2 200G + F ___________
π _ 4 (4 5 2 − 3 2 2 )100G
× 200G: ∴ 96M = 1 414,711F + 2 543,935F ∴ F = 24,251 kN
∴ σs = F __ AS = 24,251k _____
π _ 4 3 0 2 = 34,31 MPa (T)
σs = F __ A C = 24,251k ________ π _ 4 (4 5 2 − 3 2 2 )
= 30,85 MPa (C)
12.2 FF = LO + δxs
= 800 + αtL + x x = σL __ E
= 800 + (80 × 12 × 10–6 × 0,8) + 34,31M × 0,8 _________ 200G = 800 + 0,905 = 800,905 mm 12.3
C
C
SFT FT
FT = FC + FS
xs = xc = 0,5 mm
F S L ___ A S E S = 0,5 × 10–3
∴ FS = A S E S × 0,5 × 1 0 −3 ___________ L
= ( π _ 4 3 0 2 ) ( 200G ) ( 0,5 × 1 0 −3 )
_______________ 0,800905
= 88,257 kN
∴ x = F C L ____ A C E C = 0,5 × 10–3
∴ FC = π _ 4 ( 4 5 2 − 3 2 2 ) ( 100G ) 0,5 × 1 0 −3
__________________ 0,800905
= 49,081 kN
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43Module 3 • Compound bars and temperature-induced stresses
∴ F = FC + FS
= 49,081 + 88,257 = 137,338 kN
12.4 σs = F __ A S = 88,257k _____
π _ 4 3 0 2 = +124,86 MPa
σc = F __ A C = 49,081 k ________ π _ 4 (4 5 2 + 3 2 2 )
= +62,43 MPa
∴ Res σc = +62,43 + 30,85 = 93,28 MPa (C) Res σs = +124,86 + 34,31 = 90,55 MPa (C)13.
xT = 0,3
∆t = t °C
FC = 0
AS = 110 mm2
AC = 210 mm2
13.1 No load in copper means there is NO stress in the copper. After a change in temperature, the resultant stress in the copper is zero, which is equal to free expansion (free expansion = no stress).
∴ ∆x = αtL
∴ ∆x = xT = 0,3 = αtL
∴ ∆t = 0,3 × 1 0 −3 _______ α c L
= 0,3 × 1 0 −3 __________ 18 × 1 0 −6 × 0,12
= 138,89 °C
13.2 Change in length of steel due to temperature = αstLs
∴ ∆L = 12 × 10–6 × 138,89 × 0,12
= 0,2 mm
∴ Distance that F must compress steel
= 0,3 – 0,2
x = 0,1 mm
∴ Force required:
x = FL ___ AE = 0,1 × 10–3
∴ F = xAE ___ L
= 0,1 × 1 0 −3 × 110 × 1 0 −6 × 200G ___________________ 0,12
= 183,33 kN
CSL = 120
F
F
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• Knowledge of diff erent rivets• Ability to calculate force and the diff erent areas, cross-sectional area and
shear area• Knowledge of Module 1
Learning outcomes
When students have completed this module, they should be able to:• Describe a thin cylinder• Calculate the tensile and longitudinal stresses in thin cylinders• Calculate the safe internal pressure for a thin cylinder• Calculate the three failing criteria for a riveted joint• Calculate the joint effi ciency for a riveted joint• Identify diff erent riveted joints
Guidelines for students
• Make sure they know the diff erent riveted joints• Understand the principles of the module• Work from fi rst principles• Always make sketches of the joint under consideration
Thin cylinders and riveted joints4
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45Module 4 • Thin cylinders and riveted joints
Exercise 4.1 SB page 106
1. Ød = 2 m
P = 1,2 MPa
t = 25 mm
1.1 Tensile stress σt = PiD ___ 2t
= 1,2M × 2 ______ 2 × 0,025
= 48 MPa
1.2 Longitudinal stress σL = PiD ___ 4t
= 1,2M × 2 ______ 4 × 0,025
= 24 MPa
2. Pi 4 MPa D = 2 m Yield stress = 500 MPa
FOS
∴ Safe stress = 500 ___ 5 = 100 MPa
∴ Tensile stress: σt = PiD ___ 2t
∴ t = 4M × 2 ______ 2 × 100M
= 40 mm
Longitudinal stress σL = PiD ___ 4t
∴ t = 4M × 2 ______ 4 × 100M
= 0,02 m
= 20 mm
Use 40 mm plate thickness.
3. 2 MPa σallowable 80 MPa
d = 1,2 m of 72%
Spherical shell; no tensile stress only longitudinal stress.
∴ σL = PiD ___ 4tη
∴ t = 4M × 1,2 __________ 4 × 80M × 0,72
= 10,42 mm
4 m
Two stresses∴ Calculate a plate thickness for each.
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46 N5 Strength of materials and structures Lecturer Guide
4. D = 1,6 m t = 12 mm
ηL = 72% ηC = 68%
70 MPa
σt = PiD ___ 2t η L
∴ Pi = 70M × 2 × 0,012 × 0,72 ______________ 1,6
= 756 kPa
σL = PiD ___ 4t η C
∴ Pi = 70M × 4 × 0,012 × 0,68 ______________ 1,6
= 1,428 MPa
Maximum allowable = 756 kPa 5. D = ? Yield stress = 550 MPa η = 75% t = 16 Pi = 1,8 MPa FOS = 6
∴ Allowable stress = 550 ___ 6 = 91,67 MPa
σt = PiD ___ 2tη
∴ D = 91,67M × 2 × 0,016 × 0,75 ________________ 1,8M
= 1,22 m
σC = PiD ___ 4tη
∴ D = 91,67M × 4 × 0,016 × 0,75 ________________ 1,8M
= 2,44 m
Use 1,22 m.
6. t = 12 mm
d = 115 m
6.1 σt = PiD ___ 2t = 2,1M × 1,5 _______ 2 × 0,012 = 131,25 MPa
6.2 σC = PiD ___ 4t = 2,1M × 1,5 _______ 4 × 0,012 = 65,63 MPa
6.3 F = Pi DL
= 2,1M × 1,5 × 3
= 9,45 MN
Smallest pressure is the max because the higher pressure will destroy the cylinder.
If a bigger diameter is used, the stress will be more than the allowable stress and destroy the cylinder.
L = 3 m
Pi = 2,1 MPa
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47Module 4 • Thin cylinders and riveted joints
6.4 F = Pi π _ 4 d 2
= 2,1M × π _ 4 1,52
= 3,71 MN
7. F = 14 MN
7.1 σ = F __ A = 14M ___ πDt = 14M __________ π × 2,5 × 0,025
= 71,3 MPa
7.2 σ = PiD ___ 4t
∴ Pi = 71,3M × 4 × 0,025 ___________ 2,5
= 2,85 MPa safe pressure
8. L = 2,5 m Pi = 1,2 MPa t =10
D = 1,2 m σL = 36 MPa σt = 72 MPa
8.1 Force resist longitudinal
Fresist = σt 2tL
= 72M × 2 × 0,01 × 2,5
= 3,6 MN
8.2 Force acting longitudinal joint
Facting = Pi × DL
= 1,2M × 2,5 × 1,2
= 3,6 MN
8.3 Circum. joint
Force resisting
Fresist = σc × πDt
= 36M × π × 1,2 × 0,01
= 1,357 MN
8.4 Force acting
Fact = Pi × π _ 4 d 2
= 1,2M × π _ 4 1,22
= 1,357 MN
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48 N5 Strength of materials and structures Lecturer Guide
9. d = 2 m Pi = 1,7 MPa σt = 80 MPa η = 75%
σt = PiD ___ 2t η c
∴ t = PiD ____ 2σcη
= 1,7M × 2 __________ 2 × 80M × 0,75
= 28,33 mm
10. σy = 450 MPa
t = 15 ηt = 50%
d = 2,1 Pi = 2 MPa
ηL = 75%
∴ σt = Pid ___ 2t η c = 2M × 2,1 __________ 2 × 0,015 × 0,75
= 186,67 MPa
∴ FOS = σ y __ σ t = 450 _____ 186,67 = 2,41
Take 3
σC = Pid ___ 4t η t = 2M × 2,1 _________ 4 × 0,015 × 0,5
= 140 MPa
∴ FOS = σ y __ σ L = 450 ___ 140 = 3,21
Use 4 as the FOS.
11. σ = 120 MPa
D = 2,5 t = 14
ηL = 80% ηC = 40%
11.1 σt = PiD ___ 2t η L
Pi = σ2t η L ____ D
= 120M × 2 × 0,014 × 0,8 ______________ 2,5
= 1,0752 MPa
σC = PiD ___ 4t η C
Pi = 120M × 4 × 0,014 × 0,4 ______________ 2,5
= 1,0752 MPa
11.2 1,0752 MPa
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49Module 4 • Thin cylinders and riveted joints
Exercise 4.2 SB page 122
1. t = 16 mm σt = 80 MPa σC = 190 MPa τ = 35 MPa
1.1 d = 6,05 √ _ t
= 6,05 √ ___
16 = 24,2 Use 25 mm diameter
1.2 Ft = Fs
(p – d)t σt = τη π _ 4 d2
(p – 25)16 × 80 = 35 × 2 × π _ 4 252
p – 25 = 26,845 p = 51,845 mm
1.3 Ft = (p – d)tσt
= (51,845 – 25)105 × 80
= 21,476 kN
Fs = n π _ 4 d 2τ
= 2 × π _ 4 0,0252 × 35 m
= 34,36 kN
FC = ndtσC
= 2 × 0,025 × 0,16 × 190M
= 152 kN
Joint is safe against crash FC > Ft an Fs.
1.4 F = ptσt
= 0,051845 × 0,016 × 80M
= 66,36 kN
∴ ηt = F C __ F = 21,476 _____ 66,36 = 32,36%
ηS = F S __ F = 34,36 ____ 66,35 = 51,78%
ηC = F C __ F = 152 ____ 66,35 = 229,1%
∴ Joint eff. = 32,36%
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50 N5 Strength of materials and structures Lecturer Guide
2.
2.1 d = 6,05 √ _ t
= 6,05 √ ___
18
= 26 mm
2.2 Ft = Fs
(p – d)t σt = η π _ 4 d2 τ
(p – 26)18 × 80 = 3 × π _ 4 262 × 50
p – 26 = 55,305
p = 81,31 mm
2.3 F = σ.A
= σt × pt
= 80M × 0,08131 × 0,018
= 117,09 kN
2.4 Ft = (p – d)t σt
= (81,31 – 26)18 × 80
= 79,649 kN
2.5 Fs = n π _ 4 d2τ
= 3 × π _ 4 262 × 50
= 79,639 kN
2.6 FC = ndt σC
= 3 × 26 × 18 × 180
= 252,72 kN
2.7 Ft = 79,639 kN
t = 18 mm σt = 80 MPa τ = 50 MPa σC = 180 MPa
mm can be used if the 106 of the stress is left out.
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51Module 4 • Thin cylinders and riveted joints
2.8 η = Ft __ F × 100 ___ 1
= 79,639 _____ 117,09 × 100 ___ 1
= 68,02%
3. t = 25
d = 32
σt = 440
τ = 377
σC = 660 MPa FOS = 6
3.1 Ft = Fs
(p – d)t σt __ 6 = η π _ 4 d2 × 1,75 τ _ 6
(p – 32)25 × 440 ___ 6 = 2 × π _ 4 322 × 1,75 × 377 ___ 6
∴ p – 32 = 96,47
p = 128,47 mm
3.2 F = ptσt
= 128,47 × 25 × 440 ___ 6
= 235,53 kN
Ft = (p – d)t σt
= (128,47 – 32)25 × 440 ___ 6
= 176,86 kN
Fs = n π _ 4 d2 × 1,75τ
= 2 × π _ 4 322 × 1,75 × 377 ___ 6
= 176,86 kN
FC = ndtσC
= 2 × 32 × 25 × 660 ___ 6
= 176 kN
Safe load is 176 kN.
3.3 η = 176 _____ 235,73
= 74,73%
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52 N5 Strength of materials and structures Lecturer Guide
4. t = 18
d = 25
p = 90
4.1 F = ptσt = 100 kN
η = p − d ____ p × 100 ___ 1
= 90 − 25 _____ 90 × 100 ___ 1
= 72,22%
4.2 Ft = (p – d) t σt
∴ σt = Ft _____ (p − d)t
= 100k ____________ (0,09 − 0,025)0,018
= 85,47 MPa
Fs = n π _ 4 d2 × 1,75τ
100k = 2 × π _ 4 0,0252 × 1,75τ ∴ τ = 58,21 MPa
FC = ndt σC
100k = 2 × 0,025 × 0,018 × σC
∴ σC = 111,11 MPa
FOS (T) = 500 ____ 85,47 = 5,85 = 6
FOS (C) = 800 _____ 111,11 = 7,2 = 8
FOS (S) = 400 ____ 58,21 = 6,8 = 7
Use a FOS of 8.
4.3 F = 2(p – d)t1 σt
100k = 2(0,09 – 0,025)0,012 σt
∴ σt = 64,1 MPa
5. D = 1,8 t = 20 p = 100 Pi = 1,5 MPa d = 30 L = 3,5 m
5.1 η = p − d ____ p × 100 ___ 1
= 100 − 30 ______ 100
= 70%
100 kN
12 = t1
If a given load is acting on a joint then the efficiency will always be p – d
____ p × 100 ___ 1 .
If pitch and rivet diameter are given then the joint efficiency will always be p – d
____ p × 100 ___ 1 .This also means that the force Ft = (p – d)tσt is the acting force on the joint.
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53Module 4 • Thin cylinders and riveted joints
∴ σt = Pid ___ 2tη
= 1,5M × 1,8 _________ 2 × 0,02 × 0,7
= 96,43 MPa
5.2 Ft = Fs
∴ (p – d)t σt = η π _ 4 d2 τ
∴ (100 – 30)20 × 96,43 = 2 × π _ 4 302 × τ
∴ τ = 95,49 MPa
5.3 Ft = FC
(p – d)t σt = ndtσC
(100 – 30)20 × 96,43 = 2 × 30 × 20σC
∴ σC = 112,5 MPa
6.
300
t = 12
d = 22
σt = 59
τ = 49
σC = 79
6.1 Ft = Fs
(p – d)tσt = n π _ 4 d2 1,75τ
(p – 22)12,59 = 2 π _ 4 222 × 1,75 × 49
p – 22 = 92,08
p = 114,1 mm
6.2 Ft = (p – d)tσt
= (114,1 – 22)12 × 59
= 65,21 kN
6.3 Fs = n π _ 4 d2 1,57τ
= 2 × π _ 4 222 × 1,57 × 49
= 65,19 kN
6.4 FC = ndtσC
= 2 × 22 × 12 × 79
= 41,712 kN
Ft acting force on joint and Ft = Fs = FC.
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54 N5 Strength of materials and structures Lecturer Guide
6.5 F = ptσt
= 114,1 × 12 × 59
= 80,78 kN
η = 41,712 _____ 80,78 × 100 ___ 1
= 51,62%
7. D = 2,1 m
d = 25 mm
τ = 80 MPa
σt = 100 MPa
7.1 Ft = Fs
(p – d)tσt = η π _ 4 d2 × 1,75τ
(p – 25)18 × 100 = 2 × π _ 4 252 × 1,75 × 80
p – 25 = 76,36
p = 101,36 mm
7.2 η = p − d ____ p × 100 ___ 1
= 101,36 − 25 ________ 101,36 × 100 ___ 1
= 75,33%
7.3 σt = Pid ___ 2tη
100M = Pi 2,1 ____________ 2 × 0,018 × 0,7533
∴ Pi = 1,29 MPa
7.4 Ft = FC
(p – d)t σt = ndtσC
(101,36 – 25)18 × 100 = 2 × 25 × 18 × σC
∴ σC = 152,72 MPa
8. t = 12
d = 18
σt = 80
τ = 35
σC = 190
E = 18
Because Ft is acting force on joint.
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55Module 4 • Thin cylinders and riveted joints
8.1 Ft = Fs
(p – d)tσt = n π _ 4 d2 τ
(p – 18)12 × 80 = 2 × π _ 4 182 × 35
p – 18 = 18,56
p = 36,55 mm
8.2 Ft = (p – d)t × σt
= (36,55 – 18)12 × 80
= 17,808 kN
Fs = 2 π _ 4 182 × 35
= 17,81 kN
FC = ndtσC
= 2 × 18 × 12 × 190
= 82,05 kN
Yes, safe against crushing.
8.3 F = ptσt
= 36,55 × 12 × 80
= 35,088 kN
η = 17,808 _____ 35,088 × 100 ___ 1
= 50,75%
9. D = 2 m
Pi = 800 kPa
η = 70%
σt = 80
σC = 92
τ = 70
9.1 σt = PiD ___ 2tη
80M = 800k × 2 ______ 2t 0,7
∴ t = 800k × 2 _________ 2 × 80M × 0,7
= 14,29 mm
≈ 15 mm
Ft is the acting force and for
crushing a force of 82,05 kN is needed.
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56 N5 Strength of materials and structures Lecturer Guide
9.2 FC = FS
ndtσC = n π _ 4 d2 τ
2 × d × 15 × 92 = 2 × π _ 4 d2 × 70
÷ d: ∴ d = 25,1
Say 26 mm
9.3 Ft = Fs
(p – d)tσt = n π _ 4 d2 τ
(p – 26)15 × 80 = 2 × π _ 4 262 × 70
p – 26 = 61,94
p = 87,94 mm
9.4 η = p − d ____ p × 100 ___ 1
= 87,94 − 26 _______ 87,94 × 100 ___ 1
= 70,43%
10. D = 220
t = 5
Pi = 5 MPa
10.1 σt = PiD ___ 2t
= 5M × 0,220 _______ 2 × 0,005
= 110 MPa
10.2 σL = PiD ___ 4t
= 5M × 0,220 _______ 4 × 0,005
= 55 MPa
10.3 F = σ × A
= 55M × πDt
= 55M × π × 0,22 × 0,005
= 190,07 kN
11. 200 240 kN
t = 14
d = 20
11.1 FS = n π _ 4 d2 × 1,75 τ
240k = n π _ 4 0,022 × 1,75 × 75M
Students must show all steps for calculations.
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57Module 4 • Thin cylinders and riveted joints
∴ η = 5,82
Say 6 rivets
11.2 F = (W – 3d)t1σt
240k = 2[0,200 – (3 × 0,02)]t1 × 105M
∴ t1 = 8,16 mm
∴ 8 mm
11.3 240k = 6 × 1,75 × π _ 4 0,022 τ
∴ τ = 72,76 MPa
240k = (w – d)t σc
240k = (200 – 3 × 20)14 σt
σt = 85,74 MPa
240k = 64d × tσc
∴ σc = 142,86 MPa
12. d = 21
t = 12
τ = 55 MPa
σt = 70 MPa
σc = 100 MPa
Rivet pitch: Ft = FS
σt(p – d)t = n π d 2 ___ 4 τ (single shear)
70(p – 21)12 = 2 π2 1 2 ___ 4 × 55
p – 21 = 45,36
p = 66,36
Use 67 mm pitch
Because Ft = FS ∴ shearing and tearing efficiencies are the same.
∴ ηs = ηt = p − d ____ p × 100 ___ 1
= 67 − 21 _____ 67 × 100 ___ 1
= 68,66%
ηC = F C __ F = σCdtη
____ σt pt × 100 ___ 1
= σcdη ___ σt p × 100 ___ 1
Two cover strips together must be more than plate thickness.∴ Use 8 total 16 and t = 14.
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58 N5 Strength of materials and structures Lecturer Guide
= 100 × 21 × 2 ________ 70 × 67 × 100 ___ 1
= 89,55%
13.
pi = 210 kPa D = 2,5 m
PC
13.1 Shear stress rivets = 360 MPa max stress
∴ Safe shear stress = 360 ___ FOS = 360 ___ 6 = 60 MPa
Shear force in rivets is caused by internal pressure acting at end of drum.
∴ Force acting at end of drum = Shear force in rivets
∴ pi π _ 4 D2 = n π _ 4 d2τ
÷ π _ 4 : ∴ piD2 = nd2τ
210k × 2,52 = n × 0,0162 × 60M
∴ Number of rivets = n = 85,45 rivets
∴ Use 86 rivets
Two rows of rivets ∴ 86 __ 2 = 43 rivets per row
13.2 Pitch of rivets = πD ___ 43 = π(2 500) ______ 43 = 182,651 mm
∴ Rivets/m = 1 000 _____ 182,651 = 5,475 rivets/m
We cannot make it 6 because it is taken per metre length.
∴ There will be part of a rivet per metre.
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• How to calculate a moment• Ability to calculate the reactions for a simply supported beam• Ability to draw a simple shear force diagram (N4 Engineering Science)• Ability to draw a simple bending moment diagram
Learning outcomes
When students have completed this module, they should be able to:• Calculate the shear force at any point on a beam• Draw a shear force diagram and determine the position of maximum bending
moment• Calculate the bending moment at any point on the beam• Draw a bending moment diagram and determine the point of zero bending
moment
Guidelines for students
• Make sure they understand the basic concept of the theory about shear force and bending moment diagrams
• Make sure they understand the rules and defi nitions for shear force and bending moment diagrams
• Always make decent sketches of the beams and cantilevers and their shear force and bending moment diagrams
• Make a good study of the module and underline important facts
5 Loading of beams
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60 N5 Strength of materials and structures Lecturer Guide
Exercise 5 SB page 151
1.
Determine C moments about A
‹ __
M = __
› M
6C = (20 × 2) + (10 × 9)
C = 21,67 N
Determine A moments about C
__
› M =
‹ __
M
6A + (3 × 10) = (20 × 4)
6A = 8,33 N
BM – D left
BMA – BMD = 0
∴ BMB = + 8,33 × 2 = +16,66 Nm
BM right MC = –10 × 3 = –30 Nm
2.
24
A B DC3
21,67
–11,67+16,66
SF – DN
BM – DNm
8,33
8,33
20
–30
10
10
2 2 2 2 2
A B D E FC
SF – DN
BM – D
15
–10–15
+10
10
–20–20
30 10
10
25 N25 N
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61Module 5 • Loading of beams
Moments about B
‹ __
M = __
› M
4D + (10 × 2) = (30 × 2) + (10 × 6) D = 25 N
Moments about D
__
› M =
‹ __
M
4B + (10 × 2) = (30 × 2) + (10 × 6) D = 25 N
BM left
MA = MF = 0
MB = –10 × 2
= –20 Nm
MC = +(25 × 2) – (10 × 4)
= + 10 Nm
BM right
MD = 10 × 2
= –20 Nm
ME = 0
3. 3.1 and 3.2
0
0
0
5030
8080125
50 2 23 3A B C D E
50
SF – D0 N
BM – D
–50–30
10 N/m
Total UDL = 10 × 10 = 100 N
Reaction: ∴ A = E = 50 N each
SFC = + 50 – (5 × 10) = 0
SFB = + 50 – (2 × 10) = 30 (left)
SFD = + 50 – (2 × 10) = –30 (right)
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62 N5 Strength of materials and structures Lecturer Guide
BM left
MA = ME = 0
MB = +(50 × 2) – ( 10 × 2 × 2 _ 2 ) = 100 – 20
= 80 Nm
MC = +(50 × 5) – ( 10 × 5 × 5 _ 2 ) = 250 – 125
= +125 Nm
BM right MD = (50 × 2) – ( 10 × 2 × 2 _ 2 ) = +80 Nm
4.
2 2 21A B C D
HE F G
13
41,89
24,2232,33
37,81
16,11
16,11
168,11
–9,89
–25,89
29,6619,78
–16
0
BM – DNm
SF – DN
0
4 N/m 6 N/m 8 N/m
x
Moments about A
‹ __
M = __
› M
∴ 9F = ( 4 × 2 × 2 _ 2 ) + [ 6 × 3 ( 3 × 3 _ 2 ) ] + [8 × 4 × 9]
= 8 + 81 + 288
F = 41,89 N Moments about F ∴
__ › M =
‹ __
M ∴ 9A = 4 × 2(8) + (6 × 3 × 4,5] = 64 + 81 A = 16,11 N
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63Module 5 • Loading of beams
BM left
MA = MG = 0
∴ MB = +(16,11 × 2) – ( 4 × 2 × 2 _ 2 ) = 34,22 – 8
= +24,22 Nm
∴ MC = +(16,11 × 3) – (4 × 2 × 2)
= +48,33 – 16
= +32,33 Nm
∴ MD = (16,11 × 6) – (4 × 2 × 5) – (6 × 3 × 1,5)
= 96,66 – 40 – 27
= +29,66 Nm
BM right
ME = (41,89 × 2) – (8 × 4 × 2)
= 83,78 – 64
= +19,78 Nm
∴ MF = –8 × 2 × 1
= –16 Nm
H position: x = 8,11 ___ 6 = 1,352 m from C (x = S.F. at C ________ UDL C to D )
∴ MH = (16,11 × 4,352) – (4 × 2 × 3,352) – ( 6 × 1,352 × 1,352 ____ 2 ) = 70,111 – 26,816 – 5,484
= 37,811 Nm
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64 N5 Strength of materials and structures Lecturer Guide
5.
Moments about A
‹ __
M = __
› M
∴ 8E = ( 4 × 3 × 3 _ 2 ) + (20 × 6) + 8 × 3 ( 7 + 3 _ 2 ) + 6 × 12
= 18 + 120 + 204 + 72
E = 414 ___ 8
= 51,75 N
Moments about E
__
› M =
‹ __
M
∴ 9A +(6 × 4) + (8 × 3 × 0,5) = (2 × 20) + (4 × 3 × 6,5)
8A + 24 + 12 = 40 + 78
A = 82 __ 8
= 10,25 N
Position of turning point H where S.F. = 0 (BM can be a max there):
Position of H = S.F. at A ______________ UDL between A and B
= 10,25 ____ 4
= 2,563 m from A
MA = MG = 0
‹ ___
MH = (10,25 × 2,563) – ( 4 × 2,563 × 2,563 ____ 2 ) = 26,27 – 13,14
= +13,13 Nm
BM – DNm
SF – DN
4 N/m 8 N/m
2,563
10,25
13,13 12,757,5
0
0
–1,75
–21,75–29,75
–14,5–40 –12
6
22
10,25 3 3 1 1 21,5
E = 51,75
20
2
6
A H B C D 12 F G
0
x
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65Module 5 • Loading of beams
‹ __
MB = (10,25 × 3) – (4 × 3 × 1,5)
= 30,75 – 18
= +12,75 Nm
‹ __
MC = (10,25 × 6) – (4 × 3 × 4,5)
= 61,5 – 54
= +7,5 Nm
___
› MD = (51,75 × 1) – (8 × 3 × 1,5) – (6 × 5)
= 51,75 – 36 – 30
= –14,25 Nm
ME = –(6 × 4) – (8 × 2 × 1)
= –24 – 16
= –40 Nm
MF = –6 × 2
= –12 Nm
6.
Moments about B
‹ __
M = __
› M
4E + (2 × 2 × 1) = (4 × 1) + (8 × 3) + (4 × 2 × 5)
∴ E = 16 kN
4
12A B8
C D16E F2 21
4
8
–8
–8
–8–4
–4
0
0
0
0
4 kN/m
SF – DkN
BM – DkN/m
8 kN4 kN2 kN/m
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66 N5 Strength of materials and structures Lecturer Guide
Moments about E
__
› M =
‹ __
M
4B + (4 × 2 × 1) = (8 × 1) + (4 × 3) + (2 × 2 × 5)
∴ B = 8 kN
BM = at a point
MA = ME = 0
∴ MB = –2 × 2 × 1
= –4 kN/m
∴ MC = (8 × 1) – (2 × 2 × 2)
= 0 kN/m
∴ MD = (16 × 1) – (4 × 2 × 2)
= 16 – 16
= 0 kN/m
∴ ME = – 4 × 2 × 1
= –8 kN/m
7.
4A E8
B C297,5(y – 2)
122,5
122,5
–37,5
–157,5170
187,57
–220
x = 3,0625 m140
D4 2
0
0
0
30 kN/m80 kN
80SF – D
kN
BM – DkNm
40 kN/my
All lines in a BM diagram are curved lines.
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67Module 5 • Loading of beams
Moments about A
‹ __
M = __
› M
∴ 8C = (40 × 4 × 2) + (30 × 6 × 7) + 80 × 10 C = 297,5 kN
Moments about C
__
› M =
‹ __
M ∴ 8A + (80 × 2) = (40 × 4 × 6) + (30 × 6 × 1) A = 122,5 kN
Position of E
∴ x = 122,5 ____ 40 (x = S.F. at A ________ UDL A to B )
= 3,0625 m from A
MA = MD = 0
∴ ME = (122,5 × 3,0625)
= − 40 × 3,062 5 2 ________ 2
= +187,57 kN/m
∴ MB = (122,5 × 4) – (40 × 4 × 2)
= +170
∴ MC = –(80 × 2) – (30 × 2 × 1)
= −220 kNm
Moments at inflection point (right) (BM = 0)
MIMF = 0 = 297,5(y – 2) – 80y – 30 × y × y _ 2
∴ –15y2 + 217,5y – 595 = 0
∴ y = −217,5 ± √ ___________________
217, 5 2 − 4(−15)(−595) _____________________ 2(−15)
= −217,5 ± 107,73 __________ −30
y = 3,659 m from D
80 kN
297,5 kN
30 kN/m
2y – 2
y
D
C
Inflection Point
Figure 1
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68 N5 Strength of materials and structures Lecturer Guide
8.
Moments about B
‹ __
M = __
› M
4D + (80 × 1) + (60 × 1 × 0,5) = (20 × 4 × 2) + (100 × 2) + (100 × 1 × 4,5)
∴ 4D + 110 = 160 + 200 + 450
D = 175 kN
Moments about D
__
› M =
‹ __
M
4B + (100 × 1 × 0,5) = (100 × 2) + (20 × 4 × 2) + (60 × 4,5) + 80 × 5
4B + 50 = 200 + 160 + 270 + 400
B = 245 kN
MA = ME = 0
MB = –(80 × 1) – ( 60 × 1 × 1 _ 2 ) = –110 kNm
MC = (245 × 2) – (20 × 2 × 1) – ( 60 × 1 × 2 1 _ 2 ) – (80 × 3)
= 490 – 40 – 150 – 240
= +60 kNm
100 kN
2452
175
100
0
0
0
0
–80
–140 max
SF – DkN
BM – DkNm
–35–75
–50
–110
–60
10565
80 kN
A B C D E1 2F 1
60 kN/my – 0,5
y – 1
100 kN/m20 kN/m
y
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69Module 5 • Loading of beams
MD = –100 × 1 × 0,5
= –50 kNm
y80 kN
245 kN
1 mA
B
F20 kN/m60 kN/m
(y – 1 _ 2 )
y – 1
y – 1 ___ 2
Figure 2
BM at inflection point F = 0
∴ MF = 0 = +245(y – 1) – 60 × 1(y – 0,5) – 80y –20(y – 1) ( y − 1 ) ____ 2
= 245y – 245 – 60y + 30 – 80y – 10(y2 – 2y + 1)
÷ 10 = 24,5y – 24,5 – 6y + 3 – 8y – y2 + 2y – 1
= –y2 + 12,5y – 22,5
∴ y = −12,5 ± √ _________________
12, 5 2 − 4(−1)(−22,5) ___________________ 2(−1)
= −125 ± 8,14 ________ −2
= 10,32 m and 2,18 m
∴ Inflection point 2,18 m from left.
Beam length = 6 m
∴ 10,32 m not applicable.
9.
SF – DkNm
BM – DkNm
4 kN/m2 kN/m
A B C D E F22222 8
8 max
max
zeroBM
TPTP
TP
4
–4–8
–8–4
0
0
0
0
16
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70 N5 Strength of materials and structures Lecturer Guide
Moments about E
__
› M =
‹ __
M
∴ 6B = 2 × 4 × 6
B = 8 kN
Moments about B
‹ __
M = __
› M
∴ 6E = (4 × 4 × 6)
E = 16 kN
MA = M1 = 0
MB = –2 × 2 × 1
= –4 kNm
MC = 8 × 2 – 2 × 4 × 2
= 0 kNm
MD = (16 × 2) – (4 × 4 × 2)
= 0 kNm
ME = –4 × 2 × 1
= –8 kNm
10.
E
1 kN/m
30 kN20
CB50
22
–25
37,5
–42
–28 max
5
10
10
20
00
0 0
DA 5 23
SF – DkN
E
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71Module 5 • Loading of beams
10.1 Moments about A
‹ __
M = __
› M
∴ 8B = (30 × 5) + (20 × 10) + (1 × 10 × 5)
8B = 150 + 200 + 50
B = 50 kN
Moments about B
__
› M =
‹ __
M
∴ 8A + (20 × 2) = (30 × 3) + (1 × 10 × 3)
8A + 40 = 90 + 30
A = 10 kN
10.2 MA = MC = 0
MD = +10 × 5 – 1 × 5 × 5 _ 2
= +37,5 kNm
MB = –(20 × 2) – ( 1 × 2 × 2 _ 2 ) = –42 kNm
10.3 ∴ Inflection point at E.
1 kN/m20 kN
250(y – 2)
y
E C
B
ME = 0 = 50(y – 2) – 20y – ( y − y _ 2 ) = 50y – 100 – 20y – 0,5y2
= –0,5y2 + 30y – 100
y = −30 ± √ _________________
3 0 2 − 4(−0,5)(−100) _________________ 2(−0,5)
= −300 ± 26,46 ________ −1
= 3,54 m from C
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72 N5 Strength of materials and structures Lecturer Guide
11.
11.1 MG = 0
Moments about C calculate value of P
∴ ‹ __
M = __
› M
10F + (10 × 4) + (4 × 3 × 2,5) = (10 × 2) + 5P + (4 × 5 × 7.5)
10F + 40 + 30 = 20 + 5P + 150
∴ 10F = 100 + 5P
∴ F = (10 + 0,5P)
MG = 0 = +6,875(10 + 0,5P) – (4 × 5 × 4,375) – P × 1,875
0 = 68,75 + 3,4375P – 87,5 – 1,875P
∴ 18,75 = 1,5625P
Substitute in ∴ P = 12 kN
11.2 From : reaction F = 10 + (0,5 × 12)
= 16 kN
Fup = Fdown
C + 16 = 10 + 12 + 10 + 12 + 20
= 64
10 kN
6,875
5 – x
10 kN4 kN/m 4 kN/m
SF kN
BM kNm
CA D
PH
GB3 1 2 3 5
1648
26
3230
16
40
0
0
0
–16–22
–18
–70
–48
–22
–10
E F
x
H
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73Module 5 • Loading of beams
C = 48 kN
11.3 MA = MF = 0
MB = –10 × 3 – 4 × 3 × 1,5 = –48 kNm
MC = –10 × 4 – 4 × 3 × 2,5 = –70 kNm
MD = –(10 × 6) – (4 × 3 × 4,5) + (48 × 2)
= –60 – 54 + 96
= −18 kNm
ME = 16 × 5 – 4 × 5 × 2,5
= 80 – 50
= 30 kNm
Position of H
∴ x = CF at E ___________ UDL from E to F
= 4 _ 4
= 1 m
Right: MH = (16 × 4) – 4 × 4 × 2
= +32 kNm
12.
MA = 0
MB = –3 × 2 = –6
MC = –(6 × 4) – (3 × 6)
= –42
Shear forcesFA = –3 NFB = –3 – 6 = –9 NFC = –3 – 6 = –9 N
0
0 0
–9 –9
–6
–42
0
–3–3
36
A
SF – D
BM – D
BC 4 2
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74 N5 Strength of materials and structures Lecturer Guide
13.
MA = 0
MB = –2 × 3 × 1,5
= –9 Nm
BM – D
MC = –2 × 3 × 2,5
= –15 Nm
MD = –(6 × 2) – (2 × 3 × 4,5)
= –39 Nm
ME = –(2 × 3 × 7,5) – (6 × 5) – (4 × 3 × 1,5)
= –93 Nm
Shear forcesFA = 0FB = –(2 × 3) = –6 NFC = –(2 × 3) – 6 = –12 NFD = – 6 – (2 × 3) = –12 NFE = –(2 × 3) – 6 – (3 × 4) = –24 N
4 N/m 2 N/m
0 0SF – D
BM – D
6 N
D
–24
–93
–39
–15
–9
–12–12
–6–6
C B A3 32 1E
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75Module 5 • Loading of beams
14.
MA = 0
MB = 0
MC = –9 × 3 = –27 Nm
MD = –9 × 8 – ( 4 × 5 × 5 _ 2 ) = –122 Nm
ME = –9 × 11 – (4 × 5 × 5,5) = 209 Nm
MF = –(9 × 14) – (4 × 5 × 8,5) – (9 × 3) = –323 Nm
15.
–38
F 3
9 9
3 5 3 3E D C B A
–38
–29–29
–9 –9
0
0
0
0
SF – D(N)
BM – D(Nm)
–323–209
–122
–27
4 N/m
Shear forcesFA = 0FB = –9 NFC = –9 NFD = –9 – (4 × 5) = –29 NFE = –(4 × 5) – 9 – 9 = –38 NFF = –9 – (4 × 5) – 9 = –38 N
8 kN/m
E
30 kN
35,14
11,1411,14
27,14
–8
–4
–18,86 34,28
–122,06
–18,86
SF – D (kN)
BM – D (kNm)
D C B A
00
3 12 2
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76 N5 Strength of materials and structures Lecturer Guide
Reaction at B – moments about E
∴ ME = 0 = –(30 × 3) – (8 × 3 × 6,5) + B × 7
= –90 – 156 + 7B
∴ B = 35,14 kN
MA = 0
MB = – ( 8 × 1 × 1 _ 2 ) = –4 kNm
MC = +(35,14 × 2) – ( 8 × 3 × 3 _ 2 ) = +34,28 kNm
MD = +(35,14 × 4) – (8 × 3 × 3,5) = –122,06 m
ME = (–30 × 3) – (8 × 3 × 6,5) + (35,14 × 7) = 0
16.
24 4
2 2
0
A B
2 N 4 N 2 N
C D E
0
0
0
2 2 2
–2 –2 –2 –2
–4 –4
SF – D(N)
SF from A
BM – D(Nm)
x x
Reactions at B and D.
Symmetrical length and load.
∴ Fup = Fdown
B + D = 2 + 4 + 2 = 8
∴ B = D = 4
MC = 0 = + (4 × x) – 2 × (2 + x)
0 = 4x – 4 – 2x
4 = 2x
∴ x = 2 ∴ B to D = 4 m
MA = ME = 0
MD = –(2 × 2) = –4 Nm = MB MC = +(4 × 2) – (2 × 4) = 0
Shear forcesFA = 0FB1 = –(8 × 1) = –8 kNFB2 = –(8 × 1) + 35,14 = +27,14 kNFC = –(8 × 3) + 35,14 = +11,14 kNFD = –(8 × 3) + 35,14 – 30 = –18,86 kNFE = –30 – (8 × 3) + 35,14 = –18,86 kN
0843 - Futuremanagers - N5 Strength of materials LG.indd 76 2019/03/05 12:32 PM
77Module 5 • Loading of beams
17. 17.1 Reaction A
__
› M =
‹ __
M about D
A × 15 = ( 2 × 15 × 15 __ 2 ) + (5 × 4 × 9)
= 225 + 180 A = 27 N
Reaction D Moments about A
‹ __
M = __
› M
15D = ( 2 × 15 × 15 __ 2 ) + 5 × 4 × 6 D = 23 N
MB = +(27 × 2) – ( 2 × 2 × 2 _ 2 ) = 54 – 4 = 50 Nm
17.2 MC = +(23 × 7) – ( 2 × 7 × 7 _ 2 ) = 161 – 49 = +112 Nm
17.3 ME = +27 × 4 – ( 2 × 4 × 4 _ 2 ) = 108 – 16
= +92 Nm
x = 19 __ 7 = 2,714 m (x = S.F. at E ________ UDL E to C = S + 2)
∴ MF = (27 × 6,714) – (2 × 4 × 4,714) – ( 7 × 2,714 × 2,714 ____ 2 ) = 181,278 – 37,712 – 25,78
= 117,786 Nm
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78 N5 Strength of materials and structures Lecturer Guide
17.4
2 2A B E F C
9 mD
2,714
2 N/m 2 N/m
23
SF.D(N)
BM – D(Nm)
27
27
0 0
2319
–9
11292
50
117,786 –23
5 N/m
x
4 7
MA = MD = 0
Total UDL = 20 N
∴ Load/m = 20 __ 4 = 5 N/m
18.
F-DiagF
ED
8
4
A0 B C
D EF
0
–4
–8
SF. – D
CBA
2 kN/m 4 kN/m
4 kN
2 m 2 m 2 m1 m 1 m
8 kN
8 kN 16 kN
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79Module 5 • Loading of beams
19.
A B
80 kN 100 kN
20 kN/m60 kN/m 100 kN/m
245 kN
2 m 2 m1 m
0
–80B
–140
–35
C
65
100105
D0E
–75SF-DkNGiven
1 m
175 kNC D E
20.
10 kN
3 m 3 m2 m 5 m1 m
4 kN/m 4 kN/m10 kN
A
A B
B
26
–10
–22–16
SF–DkN
C D4
16
E0 F
0
B C D E F48 16 kn
F-Diag
12 kN
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• How to calculate a moment• How to calculate the reactions for a simple supported beam• How to draw a simple shear force diagram• How to draw a simple bending moment diagram• Know the diff erent stresses• Engineering Science N4• Good understanding of Module 5
Learning outcomes
When students have completed this module, they should be able to calculate:• Th e size of a beam needed• Centroid of a section• Moment of inertia of standard beams and built-up beams• Maximum safe load for a beam• Section modules for a section• And be able to use hot rolled section tables
Guidelines for students
• Make sure they understand the basic concept of the theory about simple bending• Make sure they understand how to use the shear force and bending moment
diagrams to calculate information for beams• Always make decent sketches of the beams and cantilevers and their shear force
and bending moment diagrams• Make a good study of the module and underline important facts
Simple bending of beams6
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81Module 6 • Simple bending of beams
Exercise 6.1 SB page 188
1. OD = 500 M __ I = σ _ y
ID = 460
I = π __ 64 (0,54 – 0,464) = 8,701 × 10–4 m6
y = 500 ___ 2 = 0,25
∴ M = σI __ y = 60M × 8,701 × 1 0 −4 ____________ 0,25
= 208,823 kNm
But Mmax = ML + MW
= w L 2 ___ 8 + w L 2 ___ 8
208 823 = L 2 __ 8 [1 304 + 2 308]
L = √ ______
462,51
= 21,51 m
2.
400
480
y
3
L = 3 mW
4 kN/m
M __ I L = σ _ y = 100 ___ 200
M = σI __ y = 100 × 1 0 6 × 0,08 × 0, 4 3 ______________ 0,2 × 12
= 213,33 kNm
M = w L 2 ___ 8 + WL ___ 4
213,333 = 4 × 1 0 3 × 3 2 ________ 8 + W3 ___ 4
W = 278,44 kN
+
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82 N5 Strength of materials and structures Lecturer Guide
3.
3 m
300 N/m
D = 50 M __ I = σ _ y : M = σI __ y = 120M × π0,0 5 4 _________ 0,025 × 64 = 1 472,63 Nm
1 472,63 = w L 2 ___ 8 + w L 2 ___ 8
= 300 × 3 2 ______ 8 + w 3 2 ___ 8
w = 1,009 kN/m
4.
- y AT = A1y1 + A2y2
4.1 - y (400 × 50 + 60 × 500) = (500 × 60 × 250) + (400 × 50 × 525)
50 000 - y = 18 × 106
- y = 360 mm
h1 = - y – g = 360 – 250 = 110 mm
h2 = y2 – - y = 525 – 360 = 165 mm
4.2 Ixx = I1xx + A1h12 + I2yy + A2h2
2
= [ 0,06 × 0, 5 3 _______ 12 + 0,06 × 0,5 × 0,1 1 2 ] + [ 0,4 × 0,0 5 3 _______ 12 + 0,4 × 0,05 × 0,1652 ] = 9,88 × 10–4 + 5,487 × 10–4
= 1 536,7 × 10–6 m4
4.3 Iyy = I1 + A1h12 + I2 + A2h2
2 h1 = h2 = 0 = I1xx + I2yy
= 0,05 × 0, 4 3 _______ 12 × 0,5 × 0,0 6 3 _______ 12 = 275,67 × 10–6 m4
x
x
xx
x
y
y
y
y
y
_ y
y50
y2 = 525
y1 = 250
400
500
60
1
2
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83Module 6 • Simple bending of beams
4.4
∴ σ M __ I = σ _ y
M = WL ___ 4 = 200k × 5 ______ 4 = 250 kNm
∴ σmax = M y max ____ I xx = 250k × 0,36 _________
1 536,7 × 1 0 −6
= 58,57 MPa
σmin = M y min ____ I xx
= 250k × 0,19 _________ 1 536,7 × 1 0 −6
= 30,91 MPa
4.5
σmax = σmin ∴ xt = xc
∴ σ = 250k × 0,2 _________ 275,67 × 1 0 −6
σ = Mx ___ Iyy
σ = 181,38 MPa
4.6 Zmax = M ___ σ max = 250k _____ 58,57M
= 4,268 × 1 0 −3 m 3
Zmin = M ___ σ min = 250k _____ 30,91M
= 8,088 × 1 0 −3 m 3
5.
5.1 6R = (80 × 2) + ( 12 × 6 × 6 _ 2 ) R = 62,67 kN 6L = (80 × 4) + ( 12 × 6 × 6 _ 2 ) L = 89,33 kN
190
360 y max
y minNA
200yy
x
x 200
6
12 kN/m
80 kN
2
RL
4
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84 N5 Strength of materials and structures Lecturer Guide
5.2
Mmax = (89,33 × 2) – ( 12 × 2 × 2 _ 2 ) (at 80 kN point of inflection)
= 154,66 kNm
5.3
Ixx = 1 __ 12 [0,2 × 0,63 – 0,18 × 0,583]
= 6,7332 × 10–4 m4
M __ I = σ _ y σ = My ___ I
= 154,66 × 1 0 3 × 0,3 ___________ 6,7332 × 1 0 −4
= 68,91 MPa
5.4 σ at 150 mm ∴ y = 150
∴ σ = My ___ I = 154,66 × 1 0 3 × 0,15 ____________
6,7332 × 1 0 −4
= 34,45 MPa
6. 6.1
12 kN/m
–62,67
–14,67
89,3365,33
89,33 62,67
80 kN
kN.SF.D
2
00
42
y
y
200
180
580600
150
3
2
2080 1
20x3
x
x
x x
x
x
x2
x1
y
y
yy
y
y
y y 10
100
168,64
300
y1 = 5
y2 = 150 _ y
y3 = 290
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85Module 6 • Simple bending of beams
- y AT = A1y1 + A2y2 + A3y3
No. Area y Ay
1 10 × 80 800 5 4 000
2 20 × 300 6 000 150 900k
3 20 × 100 2 000 290 580k
Total 8 800 ΣA-max 1 484k
∴ yAT = Σ Area-moment
- y 8,8k = 1 484k
- y = 168,64 mm
- x AT = Σ Area-moments
- x 8 800 = Σ Area-moments
No. Area x Ax
1 800 60 48k
2 6 000 10 60k
3 2 000 70 140k
ΣA-max 248k
∴ - x AT = Σ A-moment
- x 8 800 = 248 × 103
- x = 28,18 mm
6.2 Ixx = I1yy + A1h12 + I2xx + A2h2
2 + I3yy + A3h32
h1 = 168,64 – 5 = 163,64 mm
h2 = 168,64 – 150 = 18,64 mm
h3 = 290 – 168,64 = 121,64 m
I1T = I1yy + A1h12 = 0,08 × 0,0 1 3 ________ 12 + 0,08 × 0,01 × 0,163642
= 2,143 × 10–5 m4
I2T = I2yy + A2h22 = 0,02 × 0, 3 3 _______ 12 + 0,02 × 0,3 × 0,018642
= 4,708 × 10–5 m4
I3T = I3yy + A3h32 = 0,1 × 0,0 2 3 _______ 12 + 0,1 × 0,02 × 0,121642
= 2,966 × 10–5
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86 N5 Strength of materials and structures Lecturer Guide
Ixx = I1T + I2T + I3T
= (2,143 + 4,708 + 2,966)10–5
= 9,817 × 1 0 −5 m 4
6.3
100
70
y yx
x
Y Y
20
h3h1
h2
10
_
x = 28,18
13
2
80
2060
300
X
X X
X
X
X
y
yy
y
h1 = 60 – 28,18 = 31,82
h2 = 28,18 – 10 = 18,18
h3 = 70 – 28,18 = 41,82
Iyy = I1xx + A1h12 + I2yy + A2h2
2 + I3xx + A3h32
I1T = Ixx + A1h12
= 0,01 × 0,0 8 3 ________ 12 + [0,01 × 0,08 × 0,031822]
= 1,2367 × 10–6 m4
I2T = Iyy + A2h22
= 0,3 × 0,0 2 3 _______ 12 + [0,3 × 0,02 × 0,018182]
= 2,183 × 1 0 −6 m 4
I3T = Ixx + A3h32
= 0,02 × 0, 1 3 _______ 12 + [0,02 × 0,1 × 0,041822]
= 5,164 × 1 0 −6 m 4
Iyy = I1T + I2T + I3T
= (1,2367 + 2,183 + 5,164)10–6
= 8,584 × 10–6 m4
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87Module 6 • Simple bending of beams
6.4
M = WL + w L 2 ___ 2 = (20 × 4) – ( 10 × 4 × 4 _ 2 ) = 160 kNm
20 kn
4
10 kN/m
M __ F = σ _ y
σt = My ___ I xx = 160 × 1 0 3 × 0,13136 _____________
98,17 × 1 0 −6
131,36
168,64 = ymax
Tensile
Compressive
x x
y
y
ymin
= 214,1 MPa
6.5 σmax = My ___ I = 160 × 103 kNm
Zmax = I xx ___ y max = 98,17 × 1 0 −6 ________ 0,16864
= 5,821 × 10–4 m3
6.6 kxx = √ __
I xx __ A T = √ ________
98,17 × 1 0 −6 ________ 8 800 × 1 0 −6
= 105,62 mm
kyy = √ __
I yg __ A T = √
________ 8,584 × 1 0 −6 ________
8 800 × 1 0 −6 = 31,23 mm
7.
WN2 3
5 m σ = 80 mPa
15
15
115
11510
1
1
2X X 230
x x
y
y
y
y
y
y
y100
y = 115 + 15 = 130 mm
Ixx = 2[I1yy + A1h12] + [I2xx + A2h2
2]
h1 = 122,5 mm h2 = 0
∴ Ixx = 2 [ 0,1 × 0,01 5 3 ________ 12 + 0,1 × 0,015 × 0,122 5 2 ] + 0,01 × 0,2 3 3 ________ 12
= 45,075 × 10–6 + 10,139 × 10–6
= 55,214 × 10–6 m4
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88 N5 Strength of materials and structures Lecturer Guide
M = σI __ y = 80M × 55,214 × 1 0 −6 _____________ 0,13
= 33,978 kNm
M = Wab ____ L
33 978 = W × 2 × 3 _______ 5 ∴ W = 28,315 kN
8.
1 m 1 m 1 m
3 kN/m 2 kn5 kN 30
Dx x
Ixx = 0,03 × D 3 ______ 12 = 2,5 × 1 0 −3 D 3
At fixed end: Mmax = (2 × 3) + (5 × 1) + (3 × 1 × 1,5)
= 15,5 kNm
8.1 M __ I = σ _ y Ixx = 2,5 × 10–3 D3
y = D __ 2
σ = 120 MPa
BM 1 m from free end: M1 = 2 × 1 = 2 kNm
∴ 2 × 1 0 3 ________ 2,5 × 1 0 −3 D 3
= 120 × 1 0 6 ______ 0,5D
∴ 2,5 × 10–3 D3 × 120 × 106 = 2 × 103 × 0,5 D (M × 0,5 D)
÷ D: ∴ 300k D2 = 1 000
D = √ __________
3,33 × 1 0 −3
= 57,74 mm
8.2 BM 2 m from free end: M2 = (2 × 2) + ( 3 × 1 × 1 _ 2 ) = 5,5 kNm
∴ M 2 __ I = σ _ y ∴ 5,5 × 1 0 3 ________ 2,5 × 1 0 −3 D 3
= 120 × 1 0 6 ______ 0,5D
2,5 × 10–3 D3 × 120 × 106 = 0,5D × 5,5 × 103 (M × 0,5 D)
÷ D: ∴ 300k D2 = 2 750
D = 95,74 mm
8.3 M3 = 15,5 kNm
M 3 __ I xx = σ _ y
∴ 300k D2 = 0,5 × 15,5 × 103 (M × 0,5 D)
D = 160,73 mm
In all three cases the formula stays the same and values only B.M. that change.
W
25 m
L
a b
Students must show all steps in calculations.
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89Module 6 • Simple bending of beams
9. 9.1
1
1
160300
200
200
160
•
•
• 570x x
Ixx = IxxT – 2[Ixx1 + A1h12]
= 0,3 × 0,5 7 3 _______ 12 – 2 [ 0,16 × 0, 2 3 _______ 12 + 0,16 × 0,2 × 0,12 5 2 ] = 4,63 × 10–3 – 1,213 × 10–3
= 3,4165 × 1 0 −3 m 4
M __ I = σ _ y ∴ M = σI __ y
= 130 × 1 0 6 × 3,4165 × 1 0 −3 × 2 __________________ 0,57
= 1,558 MNm
9.2 M __ I = σ _ y
∴ Z = M __ σ
= 1,588 × 1 0 6 _______ 130 × 1 0 6
= 0,0119876 m3
Zshaf = I _ y = π D 4 × 2 _____ 64D = π D 3 ___ 32
∴ π D 3 ___ 32 = 0,0119876 D3 = 0,1221 D = 496,11 mm
10.
200
200
350
500
200
hT
100
200
100 700
200
100
200
400
532
•
•
••x x
• •
• •
_ y
Y
y
y1
y2
y3
y y
yx
X X
x
xx
Y
31
2
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90 N5 Strength of materials and structures Lecturer Guide
10.1
No. Area A y Ay
1 0,532 × 0,7 0,3724 0,35 0,13034
2 0,2 × 0,4 –0,08 0,2 –0,016
3 0,2 × 0,2 –0,04 0,5 –0,02
AT 0,2524 ΣAmom 0,09434
- y AT = ΣAmom
- y 0,2524 = 0,09434
- y = 373,772 mm
10.2 Ixx = [IxxT + AThT2] – [Iyy2 + A2h2
2] – [Ixx3 + A3h32]
hT = - y – yT = 373,772 – 350 = 23,772
∴ IxxT = 0,532 × 0,70 0 3 _________ 12 + 0,532 × 0,7 × 0,0237722 (for total cross-section including the holes) = 15,416 × 10–3 m4
I2T = (Iyy + A2h22) h2 = 373,772 – 200 = 173,772
= 0,4 × 0, 2 3 ______ 12 + 0,4 × 0,2 × 0,1737222
= 2,682 × 10–3 m4
I3T = I3xy + A3h32 h3 = 500 – 373,772 = 126,228
= 0, 2 4 ___ 12 + 0,22 × 0,126,2282
= 7,7067 × 10–4 m4
∴ True Ixx = IxxT – I2T – I3T
∴ True: Ixx = 15,416 × 10–3 – 2,682 × 10–3 – 7,7067 × 10–4
= 11,96 × 1 0 −3 m 4
10.3 Iyy = Iyysolid – Ixx2 – Iyy3 (h = 0)
= 0,7 × 0,53 2 3 ________ 12 – 0,2 × 0, 4 3 ______ 12 – 0, 2 4 ___ 12
= 8,783 × 10–3 – 1,2 × 10–3
= 7,583 × 10–3 m4
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91Module 6 • Simple bending of beams
10.4 M = wL __ 4 + w L 2 ___ 8
= 10 × 5 ____ 4 + 12 × 5 2 _____ 8
= 50 kNm
σmax= M y max ____ Ixx
S
10 kW
12 kN/m
= 50k × 0,373772 __________ 11,96 × 1 0 −3
= 1,56 MPa
σmin = M y min ____ Ixx
= 50k × 0,326228 __________ 11,96 × 1 0 −3
= 1,36 MPa
10.5
11.
(a) Zxx = I _ y = π ( D 4 − d 4 ) × 2 _________ 64 D
= π __ 32 ( 0,0 8 4 − 0,01698 4 4 ___________ 0,08 ) = 50,163 × 10–6 m3
(b)
15
150
60 • • 60
• 20
15
1 1
2X
Y Y
X
42,560
_ y = 17,5
NA •
_ y
326,228326,228
700
ymax
ymin
373,772
Y
Y 1,36 mPa
1,56 mPa
x x x x
D = 80
d = 16,986x x
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92 N5 Strength of materials and structures Lecturer Guide
- y AT = ΣAmom
- y (2 × 15 × 60) + (150 × 20) = 2(15 × 60 × 30) + (150 × 20 × 10)
4 800 - y = 84 000
- y = 17,5 mm
Iyy = 2(I1T + A1h12) + I2 + A2h2
2
= 2 [ 0,015 × 0,0 6 3 ________ 12 + 0,015 × 0,06 × 0,012 5 2 ] + [ 0,15 × 0,0 2 3 ________ 12 + 0,15 × 0,02 × 0,007 5 2 ] = 8,2125 × 10–7 + 2,6875 × 10–7
= 1,09 × 10–6 m3
Zmax = Iyy ___ y max = 1,09 × 1 0 −6 _______ 0,0425
= 25,647 × 1 0 −6 m 3
(c) Zxx = I _ y
= B D 3 × 2 ______ 12D
= B D 2 ___ 6
= 0,06 × 0,0 8 2 ________ 6
= 64 × 1 0 −3 m 3
(d) Ixx = 1 __ 12 [BD3 – bd 3]
= 1 __ 12 [(0,06 × 0,183) – (0,04 × 0,153)]
= 1,791 × 10–5
Zxx = I xx __ y = 1,791 × 1 0 −5 ________ 0,09
= 199 × 1 0 −6 m 3
Channel = 23,647 × 10–6 m3
Pipe = 50,163 × 10–6 m3
Solid rectangular = 64 × 10–6 m3
Hollow rectangular = 199 × 10–6 m3
Y X
XX
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93Module 6 • Simple bending of beams
Exercise 6.2 SB page 204
1. - y AT = A1y1 + A2y2
TAPER
X
x
x x
x
Xh2
h1
101,
6
418,
03
734,
3
_ y =
367
,0768
3,5
341,
7
1
2
1.1 - y (17,84 × 10–3 + 1,23 × 10–3) = 17,84 × 10–3 × 0,6835 _____ 2 + 1,23 × 10–3 × 0,7343 = 7,000009 × 10–3
- y = 367,07 mm (418,03)
1.2 ∴ Ixx = Ixx + A1h12 + Ixx + A2h2
2
h1 = 734,3 – 367,07 = 367,23
h2 = 367,07 – 341,75 = 25,32
Ixx = (2,176 × 10–6 + 1,23 × 10–3 × 0,367232) + (1 363 × 10–6 + 17,34 × 10–3 × 0,025322)
= 1,681 × 10–4 + 1,374 × 10–3
= 1 542 × 10–3 m4 1 542 × 10–6 m4
Iyy =Iyy1T + Iyy2B
= 0,2528 × 10–6 + 51,83 × 10–6
= 52,083 × 1 0 −6 m 6
1.3 M __ I = σ _ y ∴ σmax = M y max ____ Ixx = 400k × 0,41803 __________ 1 542 × 1 0 −6
= 108,44 MPa
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94 N5 Strength of materials and structures Lecturer Guide
2.
y y
x x
x x
1
2
•
•h = 609,6y2 = 594,7
y1 = 304,8
25,3237,1
382,9
10,4
h2
h1 = 78,1
381 × 102 × 55,1
610 × 305 × 149
__
y
x
y
x
y
2.1 - y AT = A1y1 + A2y2
- y (19,03 × 10–3 + 7,019 × 10–3) = (19,03 × 10–3 × 0,3048) + (7,019 × 10–3 × 0,5947)
26,049 × 10–3 - y = 9,975 × 10–3
- y = 382,9 mm
2.2 Ixx = I1 + A1h12 + I2 + A2h2
2
h1 = 382,9 – 304,8 = 78,1 mm ( _ y – y1 = h1)
h2 = 594,7 – 382,9 = 211,8 mm (y2 – _ y = h2)
∴ Ixx = Ixx + A1h12 + Iyy + A2h2
2
= (1 247 × 10–6 + 19,03 × 10–3 × 0,07812) + (5,849 × 10–6 + 7,019 × 10–3 × 0,21182)
= 1,363 × 10–3 + 3,207 × 10–4
= 1 683,792 × 10–6 m4
2.3 IyyT = Ixx2 + Iyy1
= 149,1 × 10–6 + 93,08 × 10–6
= 242,18 × 1 0 −6 m 4
2.4 Zmax = I xx ___ y max = 1 683 × 1 0 −6 ________ 0,3829
= 4 397,47 × 10–6
Zmin = I xx ___ y min = 1 683 × 1 0 −6 ________ 0,2371
= 7 101,6 × 1 0 −6
2.5 M __ Iv = σv __ yv
M = σI ___ y max = 120M × 1 683,792 × 1 0 −6 ________________ 0,3829
= 527,7 kNm
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95Module 6 • Simple bending of beams
Mmax = MPL + Mweight
∴ Mmax = WL ___ 4 + W L 2 ___ 8
527,7 × 103 = W6 ___ 4 + (55,1 + 149)9,81 × 6 2 _____________ 8
W = 518 690,01 × 4 _________ 6
= 345,79 kN
3.
h = 50 __ 2 + 14,5 = 39,5
x • • • x
50 × 100
80 × 45 × 8,64 kg/m 50 kN
6 m
10 kN/mh
1
2 2
3.1 IxxT = 2Ixx2 + Ixx1
= (2 × 1,059 × 10–6) × 0,05 × 0, 1 3 _______ 12
= 6,285 × 10–6 m4
IyyT = Iyy1 + 2[Iyy2 + A2h22]
= 0,1 × 0,0 5 3 _______ 12 + 2[0,1936 × 10–6 + 1,102 × 10–3 × 0,03952] = 1,0417 × 10–6 + 3,826 × 10–6
= 4,8677 × 10–6 m4
3.2 Zxxmax = I xx ___ y max = 6,285 × 1 0 −6 ________ 0,05
= 1,257 × 1 0 −4 m 3
3.3 M = WL ___ 4 + w L 2 ___ 8
= 50 × 1 0 3 × 6 ________ 4 + 10k × 6 2 ______ 8
= 75k + 45k = 120 kNm
3.4 M __ I = σ _ y : σmax = M y max ____ I xx
= 120 × 1 0 3 × 0,05 __________ 6,285 × 1 0 −6
= 954,65 MPa
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96 N5 Strength of materials and structures Lecturer Guide
3.5 Zxx = M __ σ = 120k ________ 954,65 × 1 0 6
= 125,7 × 10–6 m3
(T.flange) 178 × 102 × 21,5 kg/m
Zexx = 170,2 × 10–6 m3 (nearest bigger value)
4.
σ = 80 MPa
7 kN/m
4 m
3 m
6 m
2 m
6 kN 3 kN
4.1 M = (3× 6) + (6 × 3) + ( 7 × 4 × 4 _ 2 ) = 18 + 18 + 56
= 92 kNm
∴ Zxx = M __ σ
= 92k ___ 80M
= 1 150 × 10–6 m3
406 × 178 × 67,2 kg/m (1 189 × 10–6 m3)
4.2 Zxx = M __ σ
∴ σ = M __ 2
= 92 × 1 0 3 ________ 1 189 × 1 0 −6
= 77,38 MPa Actual stress must be smaller than max stress. ∴ Select a section with a bigger nearest Z-value.
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97Module 6 • Simple bending of beams
5.
• •
•
•
•
y2
173,4
y1
50,8
y3
265,6
y4
70 + 206,8 + 8 + 21,7 = 306,5
y
g
x x
70
191,46
141,14
h = 206,8 _ y
306,5+
26,1= 332,6
8 3
4
2
1
5.1 - y AT = Σ A-moments
No. A × 10–3 y Ay
1 2,797 0,0508 1,421 × 10–4
2 3,8 0,1734 6,5892 × 10–4
3 2,797 0,2656 7,429 × 10–4
4 2 × 1,107 0,3065 6,7859× 10–4
AT 11,608 Σ A-mom 2,2224818 × 10–3
∴ - y AT = Σ A-mom
- y 11,608 × 10–3 = 2,2224818 × 10–3
- y = 191,46 mm
5.2 Ixx = IT1 + IT2 + IT3 + IT4
∴ IT1 = Iyy1 + A1h12 (h1 = 191,46 – 50,8 = 140,66)
= 1,135 × 10–6 + 2,797 × 10–3 × 0,140662
= 5,6474 × 1 0 −5 m 4
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98 N5 Strength of materials and structures Lecturer Guide
IT2 = Ixx2 + A2h22 (h2 = 191,46 – 173,4 = 18,06)
= 28,88 × 10–6 + 3,8 × 10–3 × 0,018062
= 3,0119 × 1 0 −5
IT3 = Iyy3 + A3h32 (h3 = 265,6 – 191,46 = 74,14)
= 1,135 × 10–6 + 2,797 × 10–3 × 0,074142
= 1,6509 × 1 0 −5
IT4 = 2[Ivv4 + A4h42] (h4 = 306,5 – 191,46 = 115,04)
= 2[0,1479 × 10–6 + 1,107 × 10–3 × 0,115042]
= 2,9596 × 10–5 m4
∴ Ixx = IT1 + IT2 + IT3 + IT4
= [5,6474 + 3,0119 + 1,6509 + 2,9596]10–5
= 132,698 × 1 0 −6 m 4
IyyT = 2Ixxchannel + Iyyi-sec + 2[Iuu + A4h42]
IT4 = 2[Iuu + A4h42] h = 86 __ 2 = 43
= 2[0,5507 × 10–6 + 1,107 × 10–3 × 0,0432]
= 5,1951 × 10–6
∴ IyyT = (2 × 13,54 × 10–6) + 3,838 × 10–6 + 5,1951 × 10–6
= 36,113 × 1 0 −6 m 4
5.3 Zmax = I xx ___ y max = 132,698 × 1 0 −6 _________ 0,19146
= 693,08 × 1 0 −6 m 3
5.4 kxx = √ __
I xx __ A T
= √ __________
132,698 × 1 0 −6 _________ 11,608 × 1 0 −3
= 106,92 mm
kyy = √ __
I yg __ A T
= √ _________
36,113 × 1 0 −6 _________ 11,608 × 1 0 −3
= 55,78 mm
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99Module 6 • Simple bending of beams
6. ρ = 1 100 kg/m3
590
•
•250
150
300
5
x
y yy2
xx x
76,87 = _ y
390
5 5
Y Y
1 12
h1
h2
h2
- y AT = Σ A-moments
- y (600 × 300) – (590 × 295) = (600 × 300 × 150) – (590 × 295 × 152,5)
5 950 - y = 457 375
- y = 76,87 mm
h1 = y1 – - y = 73,13
h2 = - y – y2 = 75,37
∴ Ixx = 2[I1 + A1h12] + I2 + A2h2
2
= 2 [ 0,005 × 0, 3 3 ________ 12 + 0,005 × 0,3 × 0,0731 3 2 ] + [ 0,59 × 0,00 5 3 ________ 12 + 0,59 × 0,005 × 0,0743 7 2 ] = 3,8544 × 10–5 + 1,6322 × 10–5
= 5,48663 × 10–5 m4
M = σIxx ___ y max ymax = 300 – 76,87 = 223,13
= 60M × 5,48663 × 1 0 −5 ______________ 0,22313
= 14,754 kNm
Weight of slime/m = Volρg
= A × L × ρg
= (0,59 × 0,25) × 1 × 1 100 × 9,81
= 1 591,6725 N/m
M = w L 2 ___ 8
∴ L = √ ___
8M ___ w
= √ ___________
8 × 14,754 × 1 0 3 __________ 1 591,6725
= 8,61 m
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100 N5 Strength of materials and structures Lecturer Guide
7. D = 200
d = 180
••
•
y
y
_ y
x x y2 176,1h2
h1
y1
6,1
276,1
134,8 73,05
141,3
↔
1
2
Channel:
h = 251,5
b = 146,1
7.1 gAT = Σ A-moment
A y Ay
1 3,992 × 10–3 73,05 2,9162 × 10–4
2 π _ 4 ( 0, 2 2 − 0,1 8 2 ) 5,969 × 10–3 176,1 1,051 × 10–3
AT 9,961 × 10–3 ΣM 1,3428 × 10–3
∴ - y AT = Σ A-moment
- y = 1,3428 × 1 0 −3 _________ 9,961 × 1 0 −3
= 134,8 mm
7.2 Ixx = I1 + A1h12 + I2 + A2h2
2
h1 = 134,8 – 73,05 = 61,75
h2 = 176,1 – 134,8 = 41,3
IT1 = I1yy + A1h12
= 4,476 × 10–6 + 3,992 × 10–3 × 0,061752
= 19,698 × 10–6 m4
IT2 = I2xx + A2h22
= π __ 64 (0,24 – 0,184) + π _ 4 (0,22 – 0,182) × 0,04132
= 2,701 × 10–5 + 1,018 × 10–5
= 37,19 × 10–6 m4
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101Module 6 • Simple bending of beams
∴ Ixx = IT1 + IT2
= 19,698 × 10–6 + 37,19 × 10–6
Iyy = 56,888 × 1 0 −6 m 4 Iyy = Ixx1 + Iyy2
= 44,28 × 10–6 + 27,01 × 10–6
Iyy = 71,29 × 10–6 m4
7.3 kxx = √ __
I xx __ A T
= √ _________
56,888 × 1 0 −6 _________ 9,961 × 1 0 −3
= 75,57 mm
kyy = √ __
I yy __ A T
= √ ________
71,29 × 1 0 −6 ________ 9,961 × 1 0 −3
= 84,6 mm Smallest k-value = 75,57 mm
7.4 Zmax = M ___ σ max
= I xx ___ y max
= 56,888 × 1 0 −6 _________ 0,1413
= 402,604 × 10–6 m3
7.5 M __ I = σ _ y ∴ M = σIxx ___ ymax
M = 175 × 1 0 6 × 56,888 × 1 0 −6 ________________ 0,1413
= 70,46 kNm
7.6 M __ I = σ _ y
∴ I = M.y ___ σ
y = ? ∴ I = ?
∴ Z = M __ σ
= 70,46 × 1 0 3 _______ 175 × 1 0 −6
= 402,63 × 10–6 m3
∴ Z = I _ y
∴ π __ 64 ( D 4 − D 4 )
________ D __ 2 = Z
∴ π __ 64 ( D 4 − d 4 )
________ D __ 2 = 402,63 × 10–6
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102 N5 Strength of materials and structures Lecturer Guide
∴ D4 – d4 _____ D = 402,63 × 10–6 × 64 ___________ 2π
∴ 0,42 – d4 _____ 0,4 = 4,101 × 10–3
0,44 – d4 = 1,64 × 10–3
d4 = 0,44 – 1,64 × 10–3
= 0,02396
∴ d = 393,4 mm
8. Solid
Hollow
BS
Ds = 2Bs
DH
dH DH = 2dH
∴ MS = MH
∴ σZS = σZH (M = σZ)
∴ ZS = ZH (σS = σH)
∴ IS __ yS = IH __ yH
yS = DS __ 2 and yH = DH __ 2
∴ BSDS3 × 2 ______ 12DS
= π(D4 – d4) × 2 _________ 64DH
∴ BSDS2
____ 6 = π(DH4 – dH
4) ________ 32DH
Substitute and in :
∴ BS(2BS)2
_____ 6 = π[(2dH)4 – dH4] _________ 32 × 2dH
∴ 4BS3
___ 6 = π(16dH4 – dH
4) _________ 32(2dH)
∴ 2BS3
___ 3 = π × 15d3H ______ 32 × 2
∴ BS3 = 3π × 15d3
H _______ 32 × 4
∴ BS3 = 1,1045 dH
3
Take 3 √__
: ∴ BS = 1,0337 dH
Percentage saving:
A H − A S _____ A H × 100 ___ 1 ∴ ( 1 − A S __ A H ) ∴ [ 1 − 2 B 2 _______
π _ 4 (4 d 2 − d 2 ) ]
∴ [ 1 − 0,849 B 2 _____ d 2
]
Substitute in ∴ [ 1 − 0,849(1,0337d ) 2 __________ d 2
] ∴ (1 – 0,9072) × 100 ___ 1
∴ 9,28%
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103Module 6 • Simple bending of beams
9.
12 kN/m2
6
6 m1,4
Pitch
σmax = 480
Acting safe stress = 450 ___ 4 = 120 MPa
Per pitch ∴ FT = W/ m 2 × A
= 12 × 103 × (1,4 × 6)
= 100,8 kN
∴ Load/m = 100,8 ____ 6
= 16,8 kN/m
∴ M = w L 2 ___ 8 = 16,8 × 6 2 ______ 8 = 75,6 kNm
∴ Z = M __ σ = 75,6 × 1 0 3 _______ 120 × 1 0 6
= 630 × 10–6 m3
Nearest 305 × 165 × 46,1 kg/m (647 × 10–6)
10.
σ _ y = E __ R ∴ y = 0,01 ___ 2 = 0,005 m and
R = 8 _ 2 + 0,01 ___ 2 = 4,005 m
∴ σ = E.y __ R
= 100 × 1 0 9 × 0,005 ___________ 4,005
10
8 mR
y = 124,84 MPa
11.
D = 250
d = 150
y
2,5 kN/m weight
21 kN
5 m
∴ I = π __ 64 (0,254 – 0,154) = 1,669 × 10–4 m4
∴ Mmax = σI __ y = 40 × 1 0 6 × 1,669 × 1 0 −4 _______________ 0,125
= 53,407 kNm
Mmax = MPL + Mweight + MUDL
∴ Mmax = WL ___ 4 + w L 2 ___ 8 + w L 2 ___ 8
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104 N5 Strength of materials and structures Lecturer Guide
∴ 53 407 = 21 × 5 ____ 4 + 2,5 × 5 2 _____ 8 + w 5 2 ___ 8
= 26 250 + 7 812,5 + w25 ___ 8
∴ w25 ___ 8 = 19 344,575
w = 6,19 kN/m (safe U.D. Load)
12.
D = 400
d = 380
15 m
y = 200d
D
12.1 I = π __ 64 (0,44 – 0,384) = 233,098 × 10–6 m4
∴ M = σI __ y = 40 × 1 0 6 × 233,098 × 1 0 −6 ________________ 0,2
= 46 619,6 Nm
∴ Mmax = w L 2 ___ 8 + w L 2 ___ 8 (Mmax = Mload + Mweight)
46 619,6 = 1 × 1 0 3 × 1 5 2 ________ 8 + w1 5 2 ____ 8
∴ 46 619,6 = 28,125 + w1 5 2 ____ 8
∴ w1 5 2 ____ 8 = 18 494,6
∴ Load/m = w = 657,59 N/m
But W = Vol ρ × g 657,59 = A × 1 × 1 200 × 9,87 Water area: ∴ A = 0,05586 m 2
12.2 Percentage area full:
Area of water ___________ Inner area of pipe × 100 ___ 1
∴ 0,05586 ______ π _ 4 × 0,3 8 2
× 100 ___ 1
= 49,25%
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• How to calculate a moment• Module 6 of this book• Th e diff erent stresses• Good knowledge of Modules 1 to 5
Learning outcomes
When students have completed this module, they should be able to:• Know the diff erence between a column and a strut• Calculate the buckling loads for columns and struts• Calculate the maximum safe load for a strut and column
Guidelines for students
• Make sure they understand the basic concept second moment of area• Make sure they understand how to use the eff ective length for columns and struts• Always make decent sketches• Make a good study of the module and underline important facts
7 Columns and struts
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106 N5 Strength of materials and structures Lecturer Guide
Exercise 7.1 SB page 218
1.
D = 210
d = 170 le = L = 3 m
1.1 PE = π 2 EI ___ le = π 2 × 210G × π __ 64 (0,21 0 4 − 0,1 7 4 )
___________________ 3 2
= 12,54 MN
1.2 PR = σ c A ______ 1 + a ( le __ R ) 2
∴ le = √ __
I __ A
= 300M × π _ 4 (0,2 1 2 − 0,1 7 2 ) _______________ 1 + 1 ____ 7 500 ( 3 _____ 0,06755 ) = √ __________
π _ 4 (0,2 1 4 − 0,1 7 4 )
__________ π _ 4 (0,2 1 2 − 0,1 7 2 )
= 2,836 MN = 0,06755 m
2. 2.1 PE = π 2 EI ___ l e 2
1,2 × 106 = π 2 × 200 × 1 0 9 I __________ 5 2
∴ Iyy = 15,198 × 10–6 m4
∴ 203 × 203 × 46,6 kg/m
le = L = 5 m
2.2 PR = σ c A ______ 1 + a ( le _ k ) 2
le = √ __
I __ A = 51,2
∴ PR = 300M × 5,882 × 1 0 −3 _____________ 1 + 1 ____ 7 500 ( 2,5 _____ 0,0512 ) 2
= 1,339 MN SR = le _ k = 2,5 _____ 0,0512
= 48,83:1
3. 3.1 PR = σ c A ______ 1 + a ( le _ k ) 2
∴ 1,5 × 106 = 80 × 1 0 6 A ________ 1 + 1 ____ 6 400 ( 2,5 __ k ) 2
∴ 1,5 × 106 = 80 × 1 0 6 A _________ 1 + 9,766 × 1 0 −4 ________
k 2
∴ 1 + 9,766 × 1 0 −4 ________ k 2
= 53,333 A
But k2 = I __ A π( D 4 − d 4 ) 4 _________ 64π( D 2 − d 2 )
= ( D 2 − d 2 )( D 2 + d 2 ) ____________ 16( D 2 − d 2 )
= D 2 + d 2 _____ 16 ∴ k2 = 0,0625 (D2 + d2)
le = 0,56 = 2,5 m
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107Module 7 • Columns and struts
Substitute in ∴ 1 + 9,766 × 1 0 −4 __________ 0,0625( D 2 + d 2 )
= 53,333 × π _ 4 (D2 – d2)
∴ 1 + 0,01562 ______ ( D 2 + d 2 )
= 41,888 (D2 – d2)
× (D2 + d2): ∴ (D2 + d2) + 0,01562 = 41,888 (D4 – d4)
∴ D2 + d2 + 0,01562 = 41,888 D4 – 41,888 d4
∴ 0,262 + d2 + 0,01562 = 41,888 (0,264) – 41,888 d4
∴ d2 + 0,08322 = 0,1914 – 41,888 d4
∴ 41,888 d4 + d2 – 0,1082 = 0
41,888 d4 + d2 – 0,1082 = 0 d4 – x2 d = x ∴ 41,888 x2 + x – 0,1082 = 0
∴ x = −1 ± √ ____________________
1 2 − 4(41,888)(−0,1082) ___________________ 2(41,888)
= − 1 ± 4,374 ________ 83,776
= 3,374 _____ 83,776
x = d2 = 0,04027 ∴ d = 200,67 mm
3.2 PE = π 2 EI ___ le = π 2 85 × 1 0 9 ( π __ 64 [ 0,2 6 4 − 0,2006 7 4 ] )
____________________ 2, 5 2
= 19,424 MN
4.
y
y y
y
_ y x x
y2 = 206,05
y1 = 101,51
2
254 × 146 × 31,6
Taper203 × 152 × 52,1
- y AT = Σ A-moments
No. A y Ay
1 6,641 × 10–3 0,101,5 6,740615 × 10–4
2 3,992 × 10–3 0,20605 8,225516 × 10–4
AT 0,010633 Σ A-mom 1,4966131 × 10–3
(D2 + d2)(D2 – d2) = (D4 – d4)
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108 N5 Strength of materials and structures Lecturer Guide
4.1 ∴ - y .0,010633 = 1,4966131 × 10–3
- y = 140,75 mm
Ixx = Ixx1 + A1h12 + Iyy2 + A2h2
2
= (47,76 × 10–6 + 6,641 × 10–3 × 0,039252) + 4,476 × 10–6 + 3,992 × 10–3 × 0,06532
= 79,46 × 10–6 m4
Iyy = Ixx2 + Iyy2
= 44,28 × 10–6 + 8,098 × 10–6
= 52,378 × 1 0 −6 m 4
4.2 S.R. k = √ __
I yg __ A T = √
_________ 52,378 × 1 0 −6 _________ 0,010633 = 0,07019 m
le _ k = 0,707 × 4,5 _______ 0,07019 = 45,33:1
4.3 PR = σA ______ 1 − a ( le _ k ) 2
= 280 × 1 0 6 × 0,010633 _____________ 1 + 1 ____ 7 500 ( 0,707 × 4,5 _______ 0,07019 ) 2
= 2,34 MN
5. 5.1 Iuu = 2(45,29 × 10–6)
= 90,58 × 10–6 m4
Ivv = 2[I + Ah2]
= 2[11,72 × 10–6 + 7,635 × 10–3 × 0,07152]
= 101,5 × 10–6 m4
u
u
x x
xxh
y
y
y
y
v
v
v
v
••
•X X
Y
Y
Ixx = Iyy = 2[I + Ah2]
= 2[28,71 × 10–6 + 7,635 × 10–3 × 0,04322]
= 85,917 × 1 0 −6 m 4
∴ Bending about xx-axis or yy-axis is the weaker axis.
∴ k = √ __
I __ A = √ ___________
85,917 × 1 0 −6 __________ 2 × 7,635 × 1 0 −3
= 0,075 m
5.2 PR = σ c A ______ 1 + a ( le _ k ) 2
2 × 106 = σ c 2 × 7,635 × 1 0 −3 ____________
1 + 1 ____ 7500 ( 0,5 × 4 _____ 0,075 ) 2
= σc 0,01395
σc = 143,39 MPa
Must check all possible axes for bending to find the smallest.
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109Module 7 • Columns and struts
5.3 σsafe = σC ___ FOS = 143,39 _____ 4 = 35,847 MPa
5.4 PE = PR
π 2 EI ___ l e
2 = σ c A ______
1 + a ( le _ k ) 2
∴ π 2 × 200G × A k 2 ___________ l e
2 = σ c A _______
1 + a ( SR ) 2
÷ A: π 2 × 200G _______ (SR ) 2 = 143,39 × 1 0 6 ________
1 + S R 2 ____ 7 500
∴ 143,39 × 106 (SR)2 = 1,974 × 1012[1 + 1,33 × 10–4(SR)2]
= 1,974 × 1012 + 2,632 × 108(SR)2
1,198 × 108(SR)2 = 1,974 × 1012
∴ SR = √ _________
1 6477,538
= 128,36
6. le = 0,707 × 6 = 4,242 PE = π 2 EI ___
l e 2
3,5 × 106 = π 2 × 200G × I _________ 4,24 2 2
I = 31,91 × 1 0 −6 m 4
w w
y y
203 × 203 × 53,5 kg/m
150 150
Iyy = Iyy × 2 [ w × 0,1 5 3 ______ 12 ] 31,91 × 10–6 = 16,78 × 10–6 + 2(2,8125 × 10–4w)
= 2(2,8125 × 10–4w)
w = 26,89 mm
7. L = 4 m D = 80 PE = 150 kN E = 200 GPa
7.1 PE = π 2 EI ___ l e
2 ∴ 150k = π 2 EI ___
4 2
∴ Iyy = 1,216 × 10–6 m4
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110 N5 Strength of materials and structures Lecturer Guide
Iyy = π __ 64 (D4 – d4) ∴ 1,216 × 10–6 m4 = π __ 64 ( 0,0 8 4 − σ _ 4 ) ∴ d4 = 0,086 – 2,477 × 10–5
= 1,619 × 10–5
d = 63,43 mm
t = D − d ____ 2 = 8,285 mm
7.2 PR = σA ______ 1 + a ( le _ k ) 2
150k = σ π _ 4 (0,0 8 2 − 0,0634 3 2 ) _____________
1 + 1 ____ 7 500 ( 4 _ k ) 2
= σ1,8667 × 1 0 −3 __________ 1 + 1 ____ 7 500 ( 4 _____ 0,0255 ) 2
k = √ __
I yg __ A
= σ1,8667 × 1 0 −3 __________ 4,275 = √ ________
1,216 × 1 0 −6 ________ 1,877 × 1 0 −3
∴ σ = 343,5 MPa = 0,0255
∴ FOS = 350 ____ 343,5 = 1,018
8.
•
•
y
y
65 h
2
2
1
20
8.1 Ixx = Ixx1 + 2[Ixx2 + A2h22]
∴ Ixx = 0,02 × 0,0 5 3 ________ 12 + 2 [ π __ 64 (0,0 8 4 − 0,0 6 4 ) + π _ 4 (0,0 8 2 − 0,0 6 2 )0,06 5 2 ] = 2,083 × 10–7 + 21,33 × 10–6
= 21,54 × 10–6 m Iyy = Iyy1 + 2Iyy2
Iyy = 0,05 × 0,0 2 3 ________ 12 + 2 [ π __ 64 (0,0 8 4 − 0,0 6 4 ) ] = 2,782 × 10–6 m
PE = π 2 EI ___ l e
2
= π 2 × 209G × 2,782 × 1 0 −6 ________________ (3,2 × 0,5 ) 2
= 2,242 MN
OD = 80ID = 60
Must check both axes to find the smallest one.
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111Module 7 • Columns and struts
σ = F __ A = 2,242 × 1 0 6 _______________________ 2 × [ π _ 4 (0,0 8 2 − 0,0 6 2 ) ] + (0,05 × 0,02)
= 415,32 MPa
8.2 S.R = le _ k
k = √ __
Iyy __ A
= √ ________
2,782 × 1 0 −6 ________ 5,398 × 1 0 −3
= 0,0227 m S.R = le _ k = 3,2 × 0,5 ______ 0,0227
= 70,48:1
9. PR = σA ______ 1 + a ( le __ k ) 2
k = √ __
Iyy __ A = √
_______
0,09 × 0,0 3 3 ________ 12 _______ 0,09 × 0,03
= √ ____
0,0 3 3 ____ 12
= 8,66 × 10–3 m
∴ 100 × 103 = 130 × 1 0 6 × 0,09 × 0,03 ______________ 1 + a 2,1 _______
8,66 × 1 0 −3
∴ 100 × 103 = (1 + 58 800a) = 351 000
1 + 58 800a = 3,51
∴ 58 800a = 2,51
a = 1 _______ 23 426,295
= 4,269 × 1 0 −5
10.
Le = L = 12 m
σLIM = 100 MPa
Iyy = 80% Ixx
x = 1 ___ 200
Ixx = 2Ixx2 + (Iyy1 + A1h1
2)
= 2(80,28 × 10–6) + 2 ( 0,3 × 0,01 5 3 ________ 12 + 0,3 × 0,015 × 0,157 5 2 ) = 383,985 × 10–6 m4
30 y
y
X X 90
y
y
x x
300
100 15
29
15
300
•
•
•
• •h
h
1
1
2 2
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112 N5 Strength of materials and structures Lecturer Guide
∴ Iyy = 0,8 × 383,985 × 10–6
= 307,188 × 10–6 m4
A = 2(0,3 × 0,15) + 2(5,876 × 10–3)
= 0,020752 m2
∴ k = √ __
Iyy __ AT
= √ _________
307,188 × 10–6 _________ 0,020752 = 0,122
10.1 PS = σCA(1 – x[SR])
= 100M × 0,020752 (1 – 1 ___ 200 [ le __ k ])
= 2 075 200 (1 – 1 ___ 200 [ 12 ____ 0,122 ])
= 2 075 200 – 1 020 590
= 1,055 MN
10.2 PR = σCA ______ 1 + a[ le __ k ]2
= 100M × 0,020752 ___________ 1 + 1 ____ 7 500 [ 12 ____ 0,122 ]2
= 906,22 kN
10.3 PE = π2EIyy ____ le2
= π2 × 200G × 307,188 × 10–6
_________________ 122
= 4,211 MN
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• Module 1 shear stress• Torque and power• Engineering Science N4• A very good knowledge of Modules 1 to 7
Learning outcomes
When students have completed this module, they should be able to:• Recognise the two limits of a shaft transmitting power and torque• Calculate the power and torque transmitted by compound shaft s• Calculate maximum safe shear stress for a shaft
Guidelines for students
• Make sure they understand the basic concept of the torque equation• Make sure they understand compound shaft s• Always make decent sketches• Make a good study of the module and underline important facts
Shafts8
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114 N5 Strength of materials and structures Lecturer Guide
Exercise 8 SB page 238
1. L = 2 m D = ? T = 20 kNm G = 80 GPa τ = 60 mPa θ = 2°
1.1 T __ J = τ __ R = Gθ __ L Shear stress: T __ J = τ 2 __ D
J = TD ___ τ 2
= 20 × 1 0 3 × D ________ 60 × 1 0 6 × 2
π __ 32 D 4 = 1,667 × 10–4 D
∴ D3 = 1,6977 × 10–3
D = 119,3 mm
Angle of twist: T __ J = Gθ __ L
J = TL __ Gθ = 20 × 1 0 3 × 2 × 188 ____________ 80 × 1 0 9 × 2 × π
∴ π __ 32 D 4 = 1,432 × 10–5
D4 = 1,459 × 10–4
D = 109,9 mm
User D = 109,9 θ = 2° and τ < 60 mPa
1.2 P = 2πNT
= 2π × 900 × 20 _________ 60
= 1 884,96 kW
= 1,885 MW
2. P = 2 MW @ 400 r/min 13% D = 100 d = 90
L = 2 m G = 80 GPa
2.1 T __ J = τ __ R = Gθ __ L
Tmax = P ___ 2πN = 2,01 0 3 × 60 _______ 2π400
= 477,465 Nm
Tmax = 477,465 × 1,13
= 539,54 Nm
∴ τ = T × R ____ J = TD __________ π __ 32 ( D 4 − d 4 ) × 2
= 539,54 × 0,1 ____________ π __ 32 (0, 1 4 − 0,0 9 4 ) × 2
= 7,99 say 8 MPa
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115Module 8 • Shafts
2.2 Angle of twist ∴ T __ J = Gθ __ L
∴ θ = TL __ JG
∴ θ = 539,54 × 2 _______________ π __ 32 (0, 1 4 − 0,0 9 4 )80 × 1 0 9
= 0,004 rad. = 0,229°
3. T __ J = τ __ R = Gθ __ L D = 2,5d
3.1 T __ J = τ __ R
Stress: J = TR __ τ = 20 × 103 × D __ 2 × 1 ______
75 × 1 0 6
π __ 32 D 4 = 1,331 × 10–4 D ∴ D3 = 1,358 × 10–3
∴ D 4 − d 4 _____ D = 1,358 × 10–3
∴ (2,5d ) 4 − d 4 ________ 2,5d = 1,358 × 10–3
15,225d 3 = 1,358 × 10–3
d 3 = 8,92 × 10–5
d = 44,68 mm ∴ D = 111,7 mm
θ of twist T __ J = Gθ __ L
J = TL __ Gθ
= 20 × 1 0 3 × 1,5 × 180 _____________ 80 × 1 0 9 × 2,1 × π
π __ 32 ( D 4 − d 4 _____ D ) = 1,023 × 10–5
∴ (2,5d ) 4 − d 4 ________ 2,5d = 1,04216 × 10–4
15,225d 3 = 1,04216 × 10–4
d 3 = 6,845 d = 19 mm and D = 47,5 mm Use d = 44,68 mm D = 111,7 mm ∴ τ = 75 MPa θ < 1,5°
IMPORTANT For Question 3.1, you must check both limits to see which is the strongest.
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116 N5 Strength of materials and structures Lecturer Guide
3.2 Tmean = T max ___ 1,12
= 20 × 1 0 3 ______ 1,12
= 17,86 kNm
P = 2πNTmean
= 2π 700 ___ 60 × 17,86 × 103
= 1,31 MW
4. D = 160 G = 85 GPa L = 3 m θ = 1,1° N =400 r/min
4.1 τ __ R = T __ I
∴ τ = RT __ J
T __ J = Gθ __ L
T = GθJ ___ L
= 85G ___ 3 × 1,1 × π _____ 180 × π0,1 6 4 _____ 3 2
= 34,998 kNm
∴ τ = RT __ J
= 0,8 × 34,998 ________ π __ 32 0,1 6 4
= 43,52 MPa 4.2 Power = 2πNT = 2π × 400 ___ 60 × 34,998 ` = 1,466 MW
4.3 PH = 1,15 PS
= 1,15 × 1,466 = 1,6859 MW
D = 2d
TH = P H ___ 2πN
= 1,6859 × 60 ________ 2π × 400
= 40,248 kNm T = π __ 16 ( D 4 − d 4 _____ D ) τ
Sub. in : ∴ 40 248 = π __ 16 ( (2d ) 4 − d 4 _______ 2d ) 43,52 × 106
7,5d 3 = 4,71 × 10–3
d 3 = 6,28 × 10–4
d = 85,63 mm D = 171,27 mm
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117Module 8 • Shafts
5. 5.1 TB = 1,45 TS TT = T1 + T2
∴ θB = θS θ1 = θ2
∴ T B L B ___ J B G B = T S L S ___ J S G S
(LB = LS)
∴ 1,45 ___________ π __ 32 ( D 4 − 0,0 7 4 ) 32
= 1 ________ π __ 32 0,0 7 4 × 80
Bronze
Steel θ 70
∴ 1,45 × 0,074 × 80 = 32(D4 – 0,074)
8,704 × 10–5 = D4 – 0,074
D4 = 6,303 × 10–5
D = 89,1 mm
5.2 TS = π __ 16 D 3 τ
= π __ 16 × 0,073 × 90 × 106
= 6,061 kNm
TH = π __ 16 ( D 4 − d 6 _____ D ) τ = π __ 16 ( 0,0891 − 0,0 7 6 _________ 0,0891 ) 55 × 106
= 4,729 kNm
(A) TH = 1,45 TS But (B) ∴ TS = T H ___ 1,45
= 6,061 × 1,45 = 4,729 ____ 1,45
= 8,788 kNm = 3,26 Nm
∴ Ttotal = TH + TS
= 4,729 + 3,26
= 7,989 kNm
5.3 Tmean = 7,989 ____ 1,12
= 7,133 kNm
P = 2πNT ____ 60 = 2π × 900 × 7,133 ___________ 60
= 672,269 kW
6. D = 120 d = ?
T = 20 kNm θ > 2°
G = 80 GPa θT = θ1 + θ2
T1 = T2
TS = TH = 20 kNm 1,2 m1 m
d = 80
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118 N5 Strength of materials and structures Lecturer Guide
π __ 16 D 3 τ = π __ 16 D 4 − d 4 _____ D τ
803 = 12 0 4 − d 4 ______ 120
61,44 × 106 = 1204 – d4
d 4 = 1204 – 61,44 × 106
= 1,4592 × 108
d = 110 mm θT = θ1 + θ2
2 × π ____ 180 = T H L H
____ J H G H = T S L S ___ J S G S
2 × π ____ 180 = 20k × 1,2 ______ J H 80G = 20k × 1 _____ J S 80G
∴ × 80G and ÷ 20k
∴ 139 626,34 = 1,2 __ J H + 1 __ J S
139 626,34 = 1,2 _________ π __ 32 (0,1 2 4 − d 4 )
+ 1 _____ π __ 32 0,0 8 4
–109 053,2 = 1,2 × 32 ________ π(0,1 2 4 − d 4 )
–8 921,89 = 1 _______ 0,1 2 4 − d 4
∴ 0,124 – d 4 = 1 _______ − 8 921,89
= –1,121 × 10–4
∴ d 4 = 0,124 + 1,121 × 10–4
d 4 = 3,194 × 10–4
d = 133,69 mm
7. d = 0,66 D
∴ VS = VH (LS = LH) ∴ AS = AH
∴ D12 = D2 – d 2
Substitute in : ∴ DS
2 = D2 – (0,66D)2
= 0,5644 D2H
Take √ __
: ∴ DS = 0,7513 DH
TS : TH
∴ DS3 : D H 4 − d H 4
_______ D H
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119Module 8 • Shafts
Substitute and in :
(0,7513 DH)3 : D H 4 − (0,66D ) 4 __________ D H
0,424 DH3 : 0,8103 DH
3
∴ 1 : 1,911
8. T = π __ 16 d 3 τ
8.1 = π __ 16 × 0,133 × 69 × 106
= 29,765 kNm
P = 2πNT = 2π × 140 ___ 60 × 29,765
= 436,381 kW 8.2 TS = TH
∴ 29 765 = π __ 16 ( D 4 − d 4 _____ D ) 84 × 106
1,805 × 10–3 = 0,1 3 4 − d 4 _______ 0,13
2,346 × 10–4 = 0,134 – d 4
d 4 = 5,1 × 10–5
d = 84,51 mm
9. 9.1 Solid T __ θ = TGJ ___ TL
= 30 × 1 0 9 × π0,00 8 4 ____________ 0,22 × 32
= 54,84 Nm/rad.
Hollow T __ θ = TGJ ___ TL
= 40 × 1 0 9 × π(0,01 2 4 − 0,0 1 4 ) _________________ 0,22 × 32
= 191,64 Nm/rad.
9.2 θS = θH TT = T1 + T2
But θS = T S __ θ S = 54,84 θ1 = θ2
∴ θS = T S ____ 54,84
T H __ θ H = 191,64
Bronze
Al
∴ θH = T H _____ 191,64
Substitute and in
∴ T S ____ 54,84 = T H _____ 191,64
∴ TH = 3,495 TS
TT = TH + TS
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120 N5 Strength of materials and structures Lecturer Guide
∴ 15 = 3,495 TS + TS
TS = 3,34 Nm
∴ TH = 11,66 Nm
9.3 TH = π __ 16 ( D 4 − d 4 _____ D ) τ 11,66 = π __ 16 ( 0,01 2 4 − 0,0 1 4 _________ 0,012 ) τ τHd = 66,38 MPa TS = π __ 16 D 3 τ 3,34 = π __ 16 0,00 8 3 τ ∴ τsd = 33,22 MPa
9.4 θSd = T S ____ 54,84
= 3,34 ____ 54,84 = 0,061 rad.
θHd = 11,66 _____ 191,64 = 0,061 rad.
0,061 × 180 ___ π =3,5°10. Gs = 2,5 GH TT = T1 + T2
TH = 3TS θ = θ
10.1 θS = θH
∴ T S L S ___ J S G S = T H L H
____ J H G H
θ 60
÷ TS × GH: ∴ T S _____ J S 2,5 G H = 3 T S ____ J H G H
1 ________ π __ 32 0,0 6 4 × 2,5 = 3 _________
π __ 32 ( D 4 − 0,0 6 4 )
3 × 0,064 × 2,5 = D4 – 0,064
D4 = 1,1016 × 10–4
D = 102,45 mm
10.2 TH = π __ 16 ( D 4 − d 4 _____ D ) τ = π __ 16 ( 0,1024 5 4 − 0,0 6 4 ___________ 0,10245 ) 49M
= 9,129 kNm
TS = π __ 16 D 3 τ
= π __ 16 × 0,063 × 84M
= 3,563 kNm
TH = 3TS
= 3 × 3,563 = 10,689
10,689 > 9,129
Value of the ratio TH is bigger than the allowable value of 9,129 kNm.
Therefore, 10,689 kNm cannot be used and will damage the shaft.
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121Module 8 • Shafts
∴ TS = T H __ 3 = 9,129 ____ 3
= 3,043 kNm
∴ TT = TH + TS
= 9,129 + 3,043
= 12,172 kNm
10.3 P = 2πNT ____ 60
= 2π × 400 ___ 60 × 12,172
= 509,86 kW
11. d = 30 D = 60 mm T __ S = τ __ R = Gθ __ L
11.1 T = π __ 16 ( D 4 − d 4 _____ D ) τ ∴ 5 × 103 = π __ 16 ( 0,0 6 4 − 0,0 3 4 ________ 0,06 ) τ ∴ τ = 125,75 MPa
11.2 Shear strain δ
δ = θR __ L
= 1 × π × 0,03 ________ 180 × 1
= 5,236 × 10–4
11.3 T __ J = Gθ __ L = τ __ R
Any one
∴ G = τL __ Rθ
= 125,75 × 1 0 6 × 1 × 180 ______________ 0,03 × 1 × π
= 240,16 GPa
11.4 Tmax = 5 __ 1,2 = 4,167 kNm
∴ P = 2πNT
= 2π 200 × 4,167 _________ 60
= 87,27 kW
12. P = 7,5 MW @ 100 r/min D = 2d d = 200
θ = 3° 80 GPa
12.1 P = 2πNT
T = 7,5 × 1 0 6 × 60 _________ 2π × 100
= 509,86 kW
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122 N5 Strength of materials and structures Lecturer Guide
T __ J = τ __ R = 716,197 MNm
∴ T = π __ 16 ( D 4 − d 4 _____ D ) τ 716,197 = π __ 16 ( 0, 4 4 − 0, 2 4 _______ 0,4 ) τ = 60,79 MPa
12.2 T __ J = τ __ R = Gθ __ L
Any one in this case:
∴ Gθ __ L = τ __ R
∴ L = GθR ___ τ
= 80 × 1 0 9 × 3 × π × 0,2 ______________ 60,79 × 1 0 6 × 180
= 13,78 m
OR Gθ __ L = T __ J
L = GθJ ___ T
= 80 × 1 0 9 × π × 3π( D 4 − d 4 ) _________________ 716 197 × 180 × 32
= 13,78 m
13. dS = 30 mm D = 45 mm ValS = ValH A S Lρg = A H Lρg ∴ AS = AH
∴ dS2 = D2 – dH
2
∴ 302 = 452 – dH2
∴ dH2 = 1 125
dH = 33,54 m
C H __ C S = G J H
___ L × L ___ G J S
= J H __ J S
= D 4 − d 4 _____ D 4 S
= 4 5 4 − 33,5 4 4 ________ 3 0 4
∴ CS = 0,286 CH
14. T = P ___ 2πN = 500k × 60 ______ 2π120
= 39,789 kNm
14.1 Tmax = 39,789 × 1,14
= 45,359 kNm
T __ J = τ __ R = Gθ __ L
All values are available, so any combination of the bending equation can be used to find L.
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123Module 8 • Shafts
D = 2d
Stress: T = Jτ __ R
45 359 = J × 55 × 1 0 6 × 2 __________ D
D = 2 425,0891 J
= 2 425,0891 × π __ 32 (D4 – d 4)
∴ 1 = 238,083 ( D 4 − d 4 _____ D ) ÷ 238,083 ∴ 4,2002 × 10–3 = ( 2 d 4 − d 4 ______ 2d ) = 7,5 d3
d 3 = 5,6 × 10–4
d = 82,43 mm D = 164,86 mm
T __ J = Gθ __ L
J = TL __ Gθ
= 45,359 × 4 × 180 ___________ 80 × 1 0 9 × 1,1 × π
π __ 32 ( D 4 − d 4 ) = 1,181 × 10–4
(2d)4 – d 4 = 1,203 × 10–3
d 4 = 8,022 × 10–5
d = 94,64 mm D = 189,28 mm (usae this shaft)
14.2 δ = θR __ L
= 1,1 × π × 0,18928 ___________ 180 × 4 × 2
= 4,542 × 10–4
15. T = 120 kNm @ 150 r/min
τ = 75 MPa
15.1 Tmax = 120 × 1,1 = 132 kNm T __ J = τ __ R ∴ T = π __ 16 ( D 3 ) τ
∴ 132k = π __ 16 D3 × 75 × 106
D3 = 8,964 × 10–3
D = 207,73 mm
15.2 d = 0,65 D
Must check both, as each is a limit in the shaft: τ shear stress θ angle of twist
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124 N5 Strength of materials and structures Lecturer Guide
D3 = D 4 − d 4 _____ D
∴ 8,964 × 10–3 = D 4 − 0,6 5 4 D 4 _________ D
= 0,8215 D3
D3 = 0,0109 D = 221,8 mm d = 144,17 mm
15.3 A S − A H _____ A S
= 207,7 3 2 − (221, 8 2 − 144,1 7 2 ) __________________ 207,7 3 2
= 34,16%
16. PH = 1,18 PS
2π(1,5NS)TH = (2πNSTS)1,18
∴ TH = 0,787 TS
VH = 0,75 VS
∴ AH = 0,75 AS
D2 – d2 = 0,75 × 0,142
D2 – d2 = 0,0147
∴ d2 = D2 – 0,0147
τH = τS
∴ 16 T H D _______
π( D 4 − d 4 ) = 16 T S ____
π D S 3
Substitute in : ∴ 0,787D _____ D 4 − d 4
= 1 ____ 0,1 4 3
∴ 2,1586 × 10–3 D = D4 – d 4
Square : ∴ d
4 = (D2 – 0,0147)2
= D4 – 0,0294 D2 + 2,161 × 10–4
Substitute in : ∴ 2,1586 × 10–3 D = D4 – (D4 – 0,0294 D2 + 2,161 × 10–4) = D 4 − D 4 + 0,0294 D2 – 2,161 × 10–4 ∴ 0,0294 D2 – 2,1586 × 10–3 D – 2,161 × 10–4 = 0 ÷ 0,0294 ∴ D2 – 0,0734D – 7,35 × 10–3 = 0
D = 0,0734 ± √ _______________________
0,07342 − 4(1)(−7,35 × 1 0 −3 ) ________________________ 2
= 0,0734 ± 0,1865 __________ 2 D = 129,96 mm
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125Module 8 • Shafts
From : D2 = d 2 + 0,0147
0,129962 = d 2 + 0,0147
∴ d = 46,8 mm
17. GB = 42 GPa Gal = 31 GPa D = 40 d = 32
17.1 CB = GJ __ L = 42G × π __ 32 0,0 3 4 _________ 0,5
= 6,68 kNm/rad. θ = θ
CAL = GJ __ L TT = T1 + T2
= 31G × π __ 32 (0,0 4 4 − 0,03 2 4 ) _______________ 0,5
θ 30 s
H
C = 500 = 9,2 kNm/rad.
∴ Total C = 15,88 kNm/rad.
17.2 TT = TB + Tal
600 = J B G B θ ____ L + J al G al θ ____ L
= θ[CB + Cal] ( JG __ L = C ) = θ × 15,88k
θ = 3,778 × 10–2 rad. = 2,16°
17.3 τB = RGθ ___ L = 0,015 × 42G × 0,03778 ______________ 0,5 = 47,6 MPa
τal = RGθ ___ L = 0,02 × 31G × 0,03778 _____________ 0,5 = 46,85 MPa
17.4 P = 2πNT = ωT = 82 × 600 = 49,2 kW
18. θT = θS + θH
TS = TH
θ = 2°
G = 82 GPa
T = 20 kNm
18.1 θT = TS + TH
= T L S ___ G J S + T L H
___ G J H
∴ 3 × π ____ 180 = 20k × 0,8 _________ 82G × π __ 32 0,0 8 4
+ 20k L H ______________
82G × π __ 32 (0, 1 4 − 0,0 7 4 )
∴ 0,05236 = 0,04852 + 3,8399 × 10–3 = 0,03269 LH
∴ LH = 117,45 mm
L = 800
ф 80
L
ф = 70ф 100
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126 N5 Strength of materials and structures Lecturer Guide
18.2 TS = TH ∴ C = T __ θ = 20k _____ 3 × π ___ 180
= 381,982 kNm (rad.)
19. G = 84 GPa T = T = T θT = θ1 + θ2 + θ3
1 2
3
150 250150
D ф 80ф 70D = 80
19.1 CH = C S 3
GJ __ L = GJ __ L
π __ 32 (0,0 8 4 − 0,0 7 4 )
___________ 0,15 = π __ 32 D 4 )
____ 0,25
∴ D4 = 0,25(0,0 8 4 − 0,0 7 4 ) ____________ 0,15 D = (2,825 × 10–5)0,25
= 72,9 mm
19.2 θT = θ1 + θ2 + θ3
= τ L ___ GR + τ L ___ GR + τ L ___ GR
= τ __ G [ L 1 __ R 1 + L 2 __ R 2
+ L 3 __ R 3 ]
= 60M ___ 84G [ 0,25 _____ 0,0729 + 0,15 ___ 0,04 + 0,15 ___ 0,04 ] = 0,0103 rad. = 0,588°
T3 = τ J 3 __ R 3 = 60M × π __ 32 0,072 9 4 × 2
_____________ 0,0729 = 4,564 kNm
T1 = τ J 1 __ R 1 = 60M × π __ 32 (0,0 8 4 − 0,0 7 4 ) × 2
_________________ 0,08 = 2,496 kNm
∴ Max allowable torque = 2,496 kNm T1 = T2 = T3
19.3 P = 2πNT = ωT = 20π × 2 496
= 156,828 kW
19.4 ω = 2πN ∴ 20π = 2π N
∴ N = 10 r/s
= 10 × 60 = 600 r/min
Note for Lecturers:Questions 13, 17, 18 and 19 are not in the syllabus and may be left out.
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• Solving of simultaneous equations• Equilibrium conditions• Newton’s third law• What a vector is• What the components of a force are• Engineering Science N4
Learning outcomes
When students have completed this module, they should be able to:• Solve two unknown forces acting on an object• Calculate the equilibrants and resultant forces• Perform graphical and analytical solutions for three or more forces
Guidelines for students
• Make sure they understand the basic concept of forces in equilibrium• Make sure they know the equilibrium conditions• Always make decent sketches• Make a good study of the module and underline important facts
Forces9
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128 N5 Strength of materials and structures Lecturer Guide
Exercise 9 SB page 255
1.
45°E cos θ
E sin θ
80°
80 sin 80°
15 sin 45°
15 cos 41°
80 cos 80°
40
20
θ
Σ H.C. = 0
∴ – E cos θ – 80 cos 80° + 40 + 15 cos 45° = 0
∴ – E cos θ – 13,892 + 40 + 10,607 = 0
∴ – E cos θ = –36,715 N
Σ V.C. = 0
∴ – E sin θ + 80 sin 80° – 15 sin 45° + 20 = 0
∴ – E sin θ + 78,785 – 10,607 + 20 = 0
∴ – E sin θ = –88,178 N
θ
VC
E
88,178
36,715
θ
VC
HC
∴ E = √ _______________
36,71 5 2 + 88,17 8 2
= 95,516 N
Tan θ = 88,178 _____ 36,715 ∴ θ = 67,39°
E = 95,516 N W 67,39 S
∴ Equilibrium force = 95,516 N E 67,39° S
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129Module 9 • Forces
2.
50 cos 35°
150 sin 30°
150 cos 30°
E cos θ
E sin θ50 sin
90
100
35 θ
Σ H.C. = 0 ∴ – E cos θ + 50 cos 35° + 90 – 150 cos 30° = 0 ∴ – E cos θ + 40,958 + 90 – 129,904 – E cos θ = –1,054 Σ V.C. = 0 ∴ E sin θ + 100 – 150 sin 30° + 50 sin 35° = 0 ∴ E sin θ + 100 – 75 + 28,679 = 0 ∴ – E sin θ = –53,679
θ
E
53,679
1,054
E = √ ______________
53,67 9 2 + 1,05 4 2
= 53,689 N
tan θ = 53,679 _____ 1,054 ∴ θ = 88,88°
Resultant force = E
∴ E = 53,689 W 88,88°S
3.
Q
P 56°
56°120 N
120
Q
P
∴ cos 56° = P ___ 120 ∴ P = 120 cos 56°
P = 67,1 N
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130 N5 Strength of materials and structures Lecturer Guide
∴ sin 56° = Q __ 20 ∴ Q = 120 sin 56°
Q = 99,48 N
70°
P
P
200 N
Q
200
70Q
∴ sin 70° = 200 ___ P
∴ P = 200 _____ sin 70° = 212,84 N
tan 70° = 200 ___ Q
∴ Q = 200 _____ tan 70° = 72,79 N
4. For Figure 9.21
θ
E
E cos θ
E sin θ90°
20 N
30°
30°
200
46 N
60
30°
20 N 20 sin 30° 20 cos 30°
20 sin 30°
20 cos 30°
E cos θ θ
E sin θE
46 N
∴ Σ V.C. = 0
+ E sin θ + 20 sin 30° = 0
+ E sin θ = 10 N
Σ H.C. = 0
∴ – E cos θ + 46 – 20 cos 30° = 0
∴ – E cos θ = –28,69 N
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131Module 9 • Forces
∴ E = √ ___________
28,6 9 2 + 1 0 2
= 30,38 N
θ
10
E
28,69
E
θ
10
28,69
tan θ = 10 ____ 28,69 θ = 19,22°
Reaction = 30,38 N W19,22° N
For Figure 9.22
30°
85°
62 N 62 sin 30°62 cos 30°
80 cos 85°
80 sin 85°
θ
E sin θ
E sin θE
E cos θ
80 N
Σ V.C. = 0 E sin θ + 80 sin 85° – 62 sin 30° = 0 ∴ E sin θ + 79,7 – 31 = 0 ∴ E sin θ = –48,7 N
Σ H.C. = 0 ∴ E cos θ – 80 cos 85° – 62 cos 30° = 0 ∴ E cos θ – 6,97 – 53,69 = 0 E cos θ = +60,66 N
True position
E = √ ____________
60,6 6 2 + 48, 7 2
= 77,79 N
tan θ = 48,7 ____ 60,66
∴ θ = 38,76°
θ
48,7
є
60,66
∴ E = 77,79 N E 38,76° S
85°
30°
62 N
G
80 N
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132 N5 Strength of materials and structures Lecturer Guide
5. For Figure 9.23
B A
C
P45° 70°
120 N
Q
Scale: 1 cm = 20 kN
Q = bc = 46 kN
P = ca = 96 kN
See text book supply.
b
c
P
Q
120 N
a
For Figure 9.24
Scale: 1 cm = 20 kN
120 N
B A
C
Q P
P = ac = 74 kN
Q = bc = 74 kN
Q
120
a
b
c
P
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133Module 9 • Forces
6. For Figure 9.25
Scale: 1 cm = 10 N
100 N
80 N
A
B
C
D
65°F
P
F = ad = 92 N
P = cd = 25 N
b
c
da
7. For Figure 9.26
A
BC
D
EF
p
F 80 N
100 N60 N
50 N
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134 N5 Strength of materials and structures Lecturer Guide
F = fa = 116 N
P = ef = 50 N
80
60
50
100
ab
c
d
e
P
f
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• Solving of simultaneous equations• Equilibrium conditions• Newton’s third law• What a vector is• What the components of a force are• Engineering Science N4• Module 9• Module 5, taking moments to work out the reactions at supports
Learning outcomes
When students have completed this module, they should be able to:• Calculate reaction in supports of frameworks• Graphically determine the forces in members of a framework• Determine whether a member is a strut or a tie
Guidelines for students
• Make sure they understand the basic concept of forces in equilibrium• Make sure they know the equilibrium conditions• Always make decent sketches• Make a good study of the module and underline important facts
10 Structural frameworks
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136 N5 Strength of materials and structures Lecturer Guide
Exercise 10 SB page 272
Drawings are not to scale.
1. Figure 10.13
Scale: 1 cm = 1 m 1 cm = 10 N
40 40
40
C ML
B
AF
O
1
2 3
4
E
NK
d
60 60
Member Vector Force N Type
AB k1 65 S
BC l2 43 S
CD m3 43 S
DE n4 65 S
EF o4 24 T
FA o1 24 T
CF 23 40 T
FB 12 22 S
FD 34 22 S
Symmetrical loaded structure, therefore each support will be 60 N = 3 × 40 ____ 2 .
n
m
o
l3
2
14
k
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137Module 10 • Structural frameworks
Figure 10.14
Scale: 1 cm = 1 m 1 cm = 10 N
12
3
76.9 N 93,1 N
20 N
100 N 50 N
4
AF
KB
1 2 4
3
L
O
MC
E
N
D
nm
l
o
k
Member Vector Force N Type
AB k1 151 S
BC l3 98 S
CD m4 122 S
DE o4 98 T
EF o2 130 T
FA o1 130 T
FB 12 0 xx
EB 23 39 S
EC 34 23 T
Moment on A
∴ 13D = 100 × 5 + 50 × 9 + 20 × 13 D = 93,1 N
Moment on D
∴ 13A = 50 × 4 + 100 × 8 A = 76,9
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138 N5 Strength of materials and structures Lecturer Guide
Figure 10.15
Scale: 1 cm = 1 m 1 cm = 10 N
Scale: 1 cm = 10 N
n
m
l
o
k1
2
3
4
5
A B C DE
FG
K L M N30 N 20 N 40 N
47,5 NO O42,5 N 1 2 3 4 5
Member Vector Force N Type
AB k1 74 S
BC l2 55 S
CD m4 59 S
DE n5 81 S
EF o5 94 T
FG o3 63 T
GA o1 85 T
GB 12 35 S
GC 23 15 S
FC 34 8 S
FD 45 46 S
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139Module 10 • Structural frameworks
Moments about A
∴ 12E = 30 × 3 + 20 × 6 + 40 × 9 E = 47,5
Moments about E
∴ 12A = 40 × 3 + 20 × 6 + 30 × 9 A = 42,5
Figure 10.16
Space diagram Scale: 1 cm = 10 N
Vector diagram
1
p
P
AO
N
50 N
40 N
C
M
1
D
40°
60° 20°
140°
B
m
n
o
Member Vector Force N Type
AD op 112 S
DB p1 41 T
BC m1 114 T
CD n1 113 S
Reaction B = pm = 76 N Reaction A = op = 118 N In this case use only length for space diagram with correct angles. No calculations.
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140 N5 Strength of materials and structures Lecturer Guide
2.
O
E
D
40 N
58,12 N
CNM
L
K
A
B
FGH44°P
1 2 5 6
3 430 N
30 N
30 N
2,5 m
2,5 m
2 mP
2 m 2 m 2 m
Calculate reaction E moments about A ∴ 8A = 30 × 2,5 + 30 × 5 + 40 × 6 ∴ E = 58,12 N
Scale: 1 cm = 1 m 1 cm = 10 N
12
3
k
l
m
p
n
o
Member Vector Force N Type
AB l1 51 S
AH p1 77 T
BC m3 42 S
BG 32 31 S
GH p2 77 T
BH 12 0 O
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141Module 10 • Structural frameworks
3.
Scale: 1 cm = 1 m 1 cm = 10 N
Member Vector Force N Type
AB k1 87 S
BC l2 58 S
CD m4 58 S
DE n5 87 S
EF o5 34 T
FG p3 49 T
GA Q1 34 T
GB 12 40 T
GC 23 6 T
FC 34 6 T
FD 45 40 T
60 N
10 N 10 N
60 N
L
K
QP
O
N1
2 3 4
5
A
B
C
D
E
FG
M 30 N30 N
k
l
o
p
q
m
n
1
2
3
4
5
Symmetrical loaded structure
∴ Reactions is 60 N each:
10 + 10 + 30 + 40 + 30 ______________ 2 = 60
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142 N5 Strength of materials and structures Lecturer Guide
4.
40 N
60° 41°
3
3
Ao
3 3 3 3
60 N 30 N
50 N
B54,87 N
306,45
305,45
406,60
406,60
90 N
E cos θ
E sin θ
Moment on A
∴ 12B + (50 × 6) = (40 sin 60 × 3) + (40 cos 60 × 6) + (80 × 6) + (30 cos 45 × 6) + (30 sin 45 × 6)
= 103,92 + 120 + 480 + 127,28 + 127,28 12B + 300 = 958,48 B = 54,87 N
Σ V.C. = 0 ∴ E sin θ – 40 sin 60 – 60 – 30 sin 45 + 54,87 – 20 = 0 ∴ E sin θ = +34,64 + 60 + 21,21 – 54,87 + 20 = +80,98 N
Σ H.C. = 0 ∴ E cos θ + 40 cos 60 + 30 cos 45 – 50 = 0 ∴ E cos θ = –20 – 21,21 + 50 = 0 = +8,79 N
∴ E = √ ____________
80,9 8 2 + 8,7 9 2 = 81,5 N
80,98
8,79
E
80,98
θ
tan θ = 80,98 ____ 8,79 ∴ θ = 83,8°
E = 81,5 N E 83,8° N
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143Module 10 • Structural frameworks
Member Vector Force N Type
BC 01 55 S
CD 02 55 S
DE n3 105 S
EF m4 85 S
FG 45 45 S
GB p1 0 xx
GE 43 22 S
GD 32 78 T
GC 12 0 xx
5.Member Vector Force N Type
DC m1 84 T
CE n1 60 S
PE 12 20 T
FE 02 60 S
FD 23 74 S
Reaction at B = 106 N E Reaction at A = 116 N W 25,5° N
50 N
20 N60 N
C
M
DL
K
θA F
O
B
E
NP5
4
3
2 1
30 N
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144 N5 Strength of materials and structures Lecturer Guide
Scale: 1 cm = 20 N
1
2
3
45
P
k
m
l
n
o
6.
Scale: 1 cm = 1 m 1 cm = 20 N
A
KL 40 N20 N
P
60 N
30 N
MC
D
E
N
OB
1
2
12
p
mk
l
n
o
n
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145Module 10 • Structural frameworks
Member Vector Force N Type
AE kp 118 T
EC l2 82 T
CD n2 64 S
DB o1 86 S
DE 12 24 T
BE p1 40 S
Reaction at A = 118 N W Horizontal Reaction at B = 124 N E 45,5° N
7. Moments about B ∴ 6A + 30 sin 50° × 8 = 70 × 4,5 + 80 × 8 A = 315 + 640 – 183,85 = 128,53 N Reaction B = pk = 110 N W 32,5° N
Member Vector Force N Type
AB p1 70 T
BC k1 56 T
CD l2 34 T
AD n2 80 S
AC 12 74 S
Space diagram
70 N 30 N
80 NP
K
B
32,5°L
C 50°D
M
N
21
A
108,5 N
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146 N5 Strength of materials and structures Lecturer Guide
Scale: 1 cm = 20 N
Vector diagram no. 7
n
m
l
k
2
1
p
Use own length for space diagram.
8. 60 N
30 N
E
D
1
23
4
5A
C P
B
l m
k
q
n
40 N
80 N
20 N
O
107,99 N
Space diagram
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147Module 10 • Structural frameworks
Scale 1 cm = 10N
Vector diagram
k
q
o
p
m
l
3
4
5
First calculate reaction at “E” roller support. Take each length on 2 units (1 unit = 1 cm)
Moment about E ∴ __
M = __
M
∴ 4C = (30 × √ __
3 ) + (60N) + (40 × 5) + (20 × 6)
∴ 4C = 51,96 + 60 + 200 + 120
C = 107,99N
Draw vector diagram and measure length qk for reaction at E = 51,42N (E 13,5°N)
Member Vector Force N Type
AB n5 22 T
BC 45 69 S
CD 43 55 S
DB m4 46 T
CA p5 91 S
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MODULE
Pre-knowledge for this module
Knowledge assumed to be in place before starting with this module:• Where the tension part of a beam will be• Where the tension part of a cantilever will be• Know the diff erent stresses
Learning outcomes
When students have completed this module, they should be able to:• Know the diff erent foundations and what they are used for• Have general knowledge of purpose and placing of steel
Guidelines for students
• Make sure they understand the basic concept of this module• Make a good study of the module and underline important facts
11 Concrete and foundations
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149Module 11 • Concrete and foundations
Exercise 11.1 SB page 285
1. The purpose of a foundation is to distribute the load evenly over a bigger area and to prevent an uneven setting.
2. Strip foundations, used for houses and double-storey buildings built on a stable soil.
Pad foundations, used for columns and piers and reinforced with steel to strengthen and reduce the thickness of the foundation.
Piled foundations, used to transfer the weight through the soft soil to a more rigid base beneath.
Raft foundations, used for houses built on soft clay and fillings, and in severe conditions they could be used as piles.
Exercise 11.2 SB page 290
1. It can cause separation of the components and excessive shrinkage of the concrete when it dries. Water trapped in the concrete will form cavities when it dries and the water will decrease the quality of the concrete.
2. The water/cement ratio The degree of compaction How thoroughly the mixing was done How effectively the curing process was carried out The temperature at which the curing process takes place
3. Shuttering must be kept in position Cover floors with water, while water can be sprayed onto the walls Concrete can be covered by plastic sheets to keep the moisture in A waterproof film can be sprayed onto the concrete to avoid evaporation
4. Concrete is cast continuously to ensure a good strength. When silos are built, the concrete will be cast for 24 hours a day until the silo is completed. The reason for this is to prevent joints where cracks can develop if there was a stoppage in the casting process.
5. To use a good grade of aggregate with maximum allowable size. A more dense concrete is formed when the different parts fit in close together. The finer the aggregates are, the bigger the total area of all the small parts per unit mass which have to be covered by the cement paste, this will reduce the cover efficiency of the cement paste. The workability of the concrete will also be reduced.
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150 N5 Strength of materials and structures Lecturer Guide
Aggregates which are rounder and more uniform in shape mix easily and less energy is needed. Irregular-shaped aggregates are used when a concrete of ultimate strength is needed.
Additions will make the concrete more workable but will reduce the strength of the concrete.
Adding more water and cement will increase the workability, but the disadvantage of shrinkage will increase which will cause cracks. The cost per cubic metre will also be higher.
6. It must: • have a high tensile strength • bond well with the concrete • be compatible with the concrete, especially as far as the temperature
movements are concerned.
7. 7.1
7.2
8. The size of the reinforcement – the cover must be equal to the thickness of a bar and, when in groups of three or more, the cover must be equal to the diameter of a single bar of the equivalent area. The conditions to which the structure is subjected and the amount of heat it will be subjected to.
9. The reinforcement is always placed in the high tensile zone of a concrete beam. A simply supported beam has the main bars in the high tensile areas and binders are used to keep the main bars in position. A cantilever has main bars in the high tensile areas. The main bars are at the top. The main bars of a column with reinforced bars in place are placed near the outer surface to take up the tensile stress in a case of bending. The reinforcement must also be placed and maintained accurately in position. Ends passing each other should be tied together with No. 16 gauge annealed soft-iron wire.
Main bars Binders
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